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# Infinite Series

Infinite Series

Contents

Describe the terms: Sequence and Series 1

Describe Convergent Series and Divergent Series 3

Describe Oscillatory Series 5

Describe Convergence and Divergence for Geometric Series 6

Describe Positive Term Series 6

Describe Comparison Test 6

Examine the Convergence/Divergence of Infinite Series using Comparison Test 6

Describe Cauchy’s Root Test 6

Examine the Convergence/Divergence of Infinite Series using Cauchy’s Root Test 6

Describe D’Alembert’s Ratio Test 6

Examine the Convergence/Divergence of Infinite Series using D’Alembert’s Ratio Test 6

Describe Raabe’s Test 6

Examine the Convergence/Divergence of Infinite Series using Raabe’s Test 6

Describe Logarithmic Test 6

Examine the Convergence/Divergence of Infinite Series using Logarithmic Test 6

Describe Cauchy’s Integral Test 6

Examine the Convergence/Divergence of Infinite Series using Cauchy’s Integral Test 6

Describe Alternating Series 6

Explain Leibniz’s Rule 6

Describe Absolute and Conditional Convergence 6

# Describe the terms: Sequence and Series

Sequences and series are two important concepts in mathematics that deal with a collection of numbers that follow a specific pattern. A sequence is an ordered list of numbers that follow a particular pattern, while a series is the sum of the terms in a sequence.

Sequence:

A sequence is a collection of numbers arranged in a specific order. Each number in a sequence is called a term, and the order of the terms is important. The terms of a sequence can be finite or infinite, and can be defined using a formula, a recursive definition, or some other method.

There are various types of sequences, such as arithmetic sequences, geometric sequences, and Fibonacci sequences. In an arithmetic sequence, each term is obtained by adding a constant difference to the previous term. In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio.

Series:

A series is the sum of the terms of a sequence. If we have a sequence a1, a2, a3, …, an, the sum of the terms is denoted by:

Sn = a1 + a2 + a3 + … + an

The sum of the terms of an infinite sequence is denoted by:

S = a1 + a2 + a3 + …

If the series Sn or S converges to a finite value, it is said to be a convergent series, and if it does not converge, it is said to be a divergent series.

Example:

Consider the sequence of numbers {1, 4, 7, 10, 13, …}. This sequence can be defined recursively as an = a(n-1) + 3, with a1 = 1. The first few terms of this sequence are 1, 4, 7, 10, 13, and so on.

To find the sum of the first n terms of this sequence, we can use the formula for the sum of an arithmetic sequence:

Sn = (n/2)(a1 + an)

Substituting the values of a1, an, and n, we get:

Sn = (n/2)(1 + (1 + 3(n-1)))

Simplifying this expression, we get:

Sn = (n/2)(3n)

Sn = (3/2)n2

This formula gives us the sum of the first n terms of the sequence. For example, the sum of the first 5 terms of the sequence is:

S5 = (3/2)(52) = 37.5

This means that the sum of the first 5 terms of the sequence {1, 4, 7, 10, 13, …} is 37.5.

In this example, we have described a sequence and used the formula for the sum of an arithmetic sequence to find the sum of the first n terms of the sequence. This is an example of how sequences and series are used in mathematics to model and analyze patterns in data.

# Describe Convergent Series and Divergent Series

In mathematics, a series is a sum of terms of a sequence. A series can be convergent or divergent. A convergent series is a series whose sum approaches a finite value, while a divergent series is a series whose sum does not approach a finite value.

Convergent Series:

A series is convergent if the sum of the terms approaches a finite value as the number of terms increases. If we have a series a1 + a2 + a3 + … + an, and the sum of the first n terms approaches a limit S as n approaches infinity, then we say that the series is convergent, and we write:

a1 + a2 + a3 + … + an + … = S

The limit S is called the sum of the series. There are various tests to determine if a series is convergent, such as the comparison test, the ratio test, and the root test.

Divergent Series:

A series is divergent if the sum of the terms does not approach a finite value as the number of terms increases. If we have a series a1 + a2 + a3 + … + an, and the sum of the first n terms does not approach a limit as n approaches infinity, then we say that the series is divergent. In other words, the sum of a divergent series is either infinite or it does not exist.

There are various ways to identify divergent series. For example, if a series has a term that is greater than the harmonic series, which is a famous divergent series, then the series is also divergent.

Example:

Consider the geometric series 2 + 4 + 8 + 16 + … This series is defined by the formula an = 2n, where n is a positive integer. The sum of the first n terms of this series is given by the formula:

Sn = a1(1 – rn) / (1 – r)

where a1 = 2 and r = 2. Substituting these values, we get:

Sn = 2(1 – 2n) / (1 – 2)

Sn = 2(n+1) – 2

As n approaches infinity, the value of 2(n+1) becomes very large, and the value of Sn approaches infinity. Therefore, this series is divergent.

Another example is the series 1 + 2 + 3 + 4 + …, which is an arithmetic series with common difference 1. To find the sum of the first n terms of this series, we can use the formula for the sum of an arithmetic series:

Sn = (n/2)(a1 + an)

where a1 = 1 and an = n. Substituting these values, we get:

Sn = (n/2)(1 + n)

As n approaches infinity, the value of Sn becomes very large, and the series is divergent.

In these examples, we have described convergent and divergent series and used formulas to determine if a series is convergent or divergent. This is an example of how sequences and series are used in mathematics to model and analyze patterns in data.

# Describe Oscillatory Series

An oscillatory series is a series that alternates in sign and oscillates around a limit. The terms of an oscillatory series are either positive or negative and they change sign in a regular pattern. Oscillatory series are also called alternating series.

An oscillatory series can be expressed as a sum of positive and negative terms, where the signs of the terms alternate. An oscillatory series can be written in the following form:

a1 – a2 + a3 – a4 + a5 – a6 + …

where a1, a2, a3, … are positive numbers.

The oscillations in an oscillatory series can become smaller and smaller as more terms are added, until the sum approaches a limit. An oscillatory series can converge to a limit or diverge to infinity, depending on the behavior of the terms.

Conditions for Convergence of Oscillatory Series:

The Leibniz criterion is a test that can be used to determine the convergence of an oscillatory series. According to the Leibniz criterion, if the terms of an oscillatory series satisfy the following conditions, the series converges:

1. The terms of the series are decreasing in absolute value: |a1| >= |a2| >= |a3| >= …
2. The limit of the absolute value of the terms is zero: lim n -> infinity |an| = 0.

If these conditions are satisfied, the oscillatory series is said to converge absolutely.

Example:

Consider the series 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + … This is an example of an oscillatory series where the signs of the terms alternate. To determine if this series converges, we can apply the Leibniz criterion.

First, we need to show that the terms of the series are decreasing in absolute value. We can see that |1| > |1/2| > |1/3| > |1/4| > |1/5| > |1/6| > … Therefore, this condition is satisfied.

Next, we need to show that the limit of the absolute value of the terms is zero. We can use the limit comparison test to show that the limit of the absolute value of the terms is equal to the limit of the harmonic series, which is a divergent series. Therefore, the limit of the absolute value of the terms is not zero, and this condition is not satisfied.

Since the conditions for convergence are not satisfied, this series does not converge. However, we can still determine the sum of the series using the alternating series test. According to the alternating series test, if the terms of an oscillatory series decrease in absolute value and approach zero, then the series converges to a limit.

Applying the alternating series test to this series, we can see that the terms decrease in absolute value and approach zero. Therefore, the series converges to a limit, which can be calculated using the formula for the sum of an alternating series:

sum = 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + … = ln(2)

In this example, we have described oscillatory series and demonstrated how to determine their convergence or divergence using the Leibniz criterion and the alternating series test. Oscillatory series can be used to model a wide range of phenomena, such as sound waves, electromagnetic waves, and vibrations in mechanical systems.

# Describe Convergence and Divergence for Geometric Series

Geometric series is a sequence of numbers in which each term is obtained by multiplying the previous term by a fixed non-zero number called the common ratio. A geometric series can be expressed as follows:

S = a + ar + ar2 + ar3 + … = Σarn

where,

a = the first term of the series

r = the common ratio

n = the number of terms

Σ = the summation notation

Convergence and divergence are the two concepts that describe the behavior of a geometric series. The convergence of a series refers to the property that the sum of its terms approaches a finite value as the number of terms increases, while divergence refers to the property that the sum of its terms goes to infinity or does not approach a finite value.

1. Convergence of Geometric Series:

A geometric series is said to converge if the sum of its terms approaches a finite value as the number of terms increases. The necessary and sufficient condition for the convergence of a geometric series is that the absolute value of the common ratio, |r|, should be less than one. That is,

|r| < 1

Example:

Consider the geometric series 3 + 1.5 + 0.75 + 0.375 + …

The first term is 3 and the common ratio is 1/2.

Then, the sum of the first n terms of the series can be written as:

Sn = 3(1 – (1/2)n)/(1 – 1/2)

Taking the limit as n approaches infinity, we get:

lim (n→∞) Sn = 3/(1 – 1/2) = 6

Therefore, the given geometric series converges to a sum of 6.

1. Divergence of Geometric Series:

A geometric series is said to diverge if the sum of its terms does not approach a finite value as the number of terms increases. The necessary and sufficient condition for the divergence of a geometric series is that the absolute value of the common ratio, |r|, should be greater than or equal to one. That is,

|r| ≥ 1

Example:

Consider the geometric series 5 + 10 + 20 + 40 + …

The first term is 5 and the common ratio is 2.

Then, the sum of the first n terms of the series can be written as:

Sn = 5(2n – 1)

Taking the limit as n approaches infinity, we get:

lim (n→∞) Sn = ∞

Therefore, the given geometric series diverges and does not have a finite sum.

1. Convergence and Divergence of Special Geometric Series:

The convergence and divergence of special geometric series can be determined based on the common ratio (r) of the series. Here are the conditions for convergence and divergence:

1. Convergent Geometric Series:
• When -1 < r < 1, the geometric series converges.
• The sum of the series can be calculated using the formula S = a / (1 – r), where ‘a’ is the first term.
2. Divergent Geometric Series:
• When r ≤ -1 or r ≥ 1, the geometric series diverges.
• The series does not have a finite sum and extends to infinity.

Here are some examples:

1. Convergent Geometric Series:

Example 1: 1/2 + 1/4 + 1/8 + 1/16 + …

Here, the common ratio is r = 1/2, which satisfies -1 < r < 1.

Therefore, the series converges, and the sum is S = (1/2) / (1 – 1/2) = 1.

Example 2: 3/5 + 3/25 + 3/125 + 3/625 + …

Here, the common ratio is r = 3/5, which satisfies -1 < r < 1.

Therefore, the series converges, and the sum is S = (3/5) / (1 – 3/5) = 15.

1. Divergent Geometric Series:

Example 1: 2 + 4 + 8 + 16 + …

Here, the common ratio is r = 2, which is greater than or equal to 1.

Therefore, the series diverges, and it does not have a finite sum.

Example 2: -3 – 6 – 12 – 24 – …

Here, the common ratio is r = -2, which is less than or equal to -1.

Therefore, the series diverges, and it does not have a finite sum.

These examples illustrate the convergence and divergence of special geometric series based on the common ratio. It’s important to note that these conditions apply specifically to geometric series and may not hold for other types of series.

# Describe Positive Term Series

A positive term series is a sequence of numbers in which all terms are positive. It is a type of series in which each term in the sequence is greater than or equal to zero. A positive term series can be represented as:

S = a1 + a2 + a3 + … + an + …

where,

ai = the ith term of the series

The behavior of a positive term series is described by its convergence or divergence. The convergence of a series refers to the property that the sum of its terms approaches a finite value as the number of terms increases, while divergence refers to the property that the sum of its terms goes to infinity or does not approach a finite value.

1. Convergence of Positive Term Series:

A positive term series is said to converge if the sum of its terms approaches a finite value as the number of terms increases. The necessary and sufficient condition for the convergence of a positive term series is that the sum of its terms is bounded, which means that there exists a real number L such that the sum of the first n terms of the series is less than or equal to L for all n. That is,

Σ ai ≤ L

Example:

Consider the series 1 + 1/2 + 1/4 + 1/8 + …

The sum of the first n terms of the series can be written as:

Sn = 1 + 1/2 + 1/4 + … + 1/2(n-1)

Using the formula for the sum of a geometric series, we get:

Sn = (1 – 1/2n)/(1 – 1/2) = 2 – 1/2n

Taking the limit as n approaches infinity, we get:

lim(n→∞) Sn = 2

Therefore, the given positive term series converges to a sum of 2.

1. Divergence of Positive Term Series:

A positive term series is said to diverge if the sum of its terms does not approach a finite value as the number of terms increases. The necessary and sufficient condition for the divergence of a positive term series is that the sum of its terms is unbounded, which means that there does not exist a real number L such that the sum of the first n terms of the series is less than or equal to L for all n. That is,

Σ ai → ∞

Example:

Consider the series 1 + 2 + 3 + 4 + …

The sum of the first n terms of the series can be written as:

Sn = 1 + 2 + 3 + … + n

Using the formula for the sum of an arithmetic series, we get:

Sn = n(n+1)/2

Taking the limit as n approaches infinity, we get:

lim(n→∞) Sn = ∞

Therefore, the given positive term series diverges and does not have a finite sum.

1. Test for Convergence of Positive Term Series:

There are several tests for determining the convergence or divergence of a positive term series. Some of the common tests are:

(i) Comparison Test: If Σai and Σbi are positive term series such that ai ≤ bi for all i and Σbi converges, then Σai also converges.

(ii) Limit Comparison Test: If Σai and Σbi are positive term series such that lim(n→∞) ai/bi = c, where c is a finite positive

# Describe Comparison Test

The comparison test is a method to determine the convergence or divergence of a series by comparing it to another series whose behavior is already known. The comparison test is based on the idea that if the terms of a series are smaller than the terms of another series that converges, then the original series must also converge. Similarly, if the terms of a series are larger than the terms of another series that diverges, then the original series must also diverge.

There are two forms of the comparison test: the direct comparison test and the limit comparison test.

1. Direct Comparison Test:

The direct comparison test states that if the terms of a series an are smaller than the terms of another series bn that converges, then the series an also converges. Similarly, if the terms of a series an are larger than the terms of another series bn that diverges, then the series an also diverges. Formally, the direct comparison test can be stated as follows:

If 0 ≤ an ≤ b for all n and Σ bn converges, then Σ an converges.

If 0 ≤ bn ≤ an for all n and Σ bn diverges, then Σ an diverges.

Example:

Consider the series Σ(1/n2) and Σ(1/n). The terms of the series Σ(1/n2) are smaller than the terms of the series Σ(1/n) for all n. It is known that Σ(1/n) is a p-series with p = 1, which is a convergent series. Therefore, by the direct comparison test, Σ(1/n2) is also a convergent series.

1. Limit Comparison Test:

The limit comparison test is used when the direct comparison test is inconclusive, that is, when the terms of the two series being compared are not strictly greater than or less than each other. The limit comparison test is based on the idea that if the limit of the ratio of the terms of two series exists and is positive, then the two series behave the same way. Formally, the limit comparison test can be stated as follows:

If an > 0 and bn > 0 for all n and lim(n→∞) an/bn = c, where c is a positive finite number,

then either both series converge or both series diverge.

Example:

Consider the series Σ(1/n2) and Σ(1/(n2 + 1)). We can use the limit comparison test to determine their behavior. We have:

lim(n→∞) (1/n2)/(1/(n2 + 1)) = lim(n→∞) (n2 + 1)/n2 = 1

Since the limit exists and is positive, by the limit comparison test, Σ(1/n2) and Σ(1/(n2 + 1)) behave the same way. It is known that Σ(1/n2) is a convergent series. Therefore, Σ(1/(n2 + 1)) is also a convergent series.

# Examine the Convergence/Divergence of Infinite Series using Comparison Test

The comparison test is a powerful tool in determining the convergence or divergence of infinite series. The test involves comparing a given series with another known series whose convergence or divergence is already known. There are two forms of the comparison test: the direct comparison test and the limit comparison test.

The Direct Comparison Test:

The direct comparison test can be used when the terms of the series being examined are greater than or equal to zero. Suppose that for all n, an and bn are non-negative terms in two series such that an ≤ bn for all n. Then, we have the following conclusions:

• If the series ∑bn converges, then the series ∑an converges.
• If the series ∑an diverges, then the series ∑bn diverges.

In other words, if a series is bounded above by a convergent series, then it is also convergent, and if it is bounded below by a divergent series, then it is also divergent.

Example:

Consider the series ∑(1/n2 + 1) and ∑(1/n2). We can use the direct comparison test to determine if the given series converges or diverges. Since the terms of the series ∑(1/n2 + 1) are greater than the terms of the series ∑(1/n2), we can bound the series ∑(1/n2 + 1) by the series ∑(1/n2) to conclude that ∑(1/n2 + 1) is convergent. The series ∑(1/n2) is known as the p-series with p = 2, which is a convergent series.

The Limit Comparison Test:

The limit comparison test can be used when the terms of the series being examined are not strictly greater than or less than each other. Suppose that an and bn are two series of positive terms such that lim(n→∞) an/bn = c, where c is a positive constant. Then we have the following conclusions:

• If the series ∑bn converges, then the series ∑an also converges.
• If the series ∑an diverges, then the series ∑bn also diverges.

In other words, if the ratio of the terms of two series converges to a finite non-zero constant, then the behavior of the series is the same.

Example:

Consider the series ∑(1/n3 + 1) and ∑(1/n3). We can use the limit comparison test to determine if the given series converges or diverges. Since lim(n→∞) (1/n3 + 1) / (1/n3) = 1, the two series behave in the same way. The series ∑(1/n3) is known as the p-series with p = 3, which is a convergent series. Therefore, ∑(1/n3 + 1) is also convergent.

In conclusion, the comparison test is a useful tool for determining the convergence or divergence of series. When the terms of the series being examined are greater than or equal to zero, the direct comparison test is used. When the terms are not strictly greater or less than each other, the limit comparison test is used.

# Describe Cauchy’s Root Test

Cauchy’s Root Test is a method for determining the convergence or divergence of a series. It is a powerful test that can be applied to series with positive, negative or complex terms. The test is named after Augustin-Louis Cauchy, a French mathematician.

Cauchy’s Root Test:

Suppose that we have a series ∑an. We define the nth root of the absolute value of an as

rn = (|an|)(1/n).

We then take the limit as n goes to infinity:

lim(n→∞) rn = L.

The series ∑an is convergent if L < 1 and divergent if L > 1. The test is inconclusive if L = 1.

Example:

Consider the series ∑(n2 / 2n). We can use Cauchy’s Root Test to determine if the given series converges or diverges. We define the nth root of the absolute value of an as:

rn = (|n2 / 2n|)(1/n) = (n(2/n)) / (2(1/n))

Taking the limit as n goes to infinity, we have:

lim(n→∞) rn = lim(n→∞) [(n(2/n)) / (2(1/n))]

= lim(n→∞) [e(ln(n(2/n)) – ln(2(1/n)))]

= lim(n→∞) [e[(2/n) ln(n) – (1/n) ln(2)]]

= e0 = 1.

Since L = 1, the test is inconclusive. We can use another test, such as the Ratio Test, to determine the convergence or divergence of the series.

In conclusion, Cauchy’s Root Test is a useful tool for determining the convergence or divergence of series. It is particularly useful for series with positive, negative or complex terms, and can often be used when other tests fail. However, the test is inconclusive when L = 1, and another test must be used to determine the convergence or divergence of the series.

# Examine the Convergence/Divergence of Infinite Series using Cauchy’s Root Test

Cauchy’s Root Test is a powerful tool for determining the convergence or divergence of infinite series. This test is particularly useful when the terms of the series have a complicated form or when other tests fail to provide conclusive results.

To apply Cauchy’s Root Test, we need to find the nth root of the absolute value of the nth term of the series, which is given by:

rn = (|an|)(1/n)

We then take the limit of rn as n approaches infinity:

L = lim(n→∞) rn

If L < 1, then the series converges absolutely. If L > 1, then the series diverges. If L = 1, then the test is inconclusive, and we need to apply other tests to determine convergence or divergence.

Example:

Consider the series ∑ (n!) / (nn). We can use Cauchy’s Root Test to determine if the given series converges or diverges.

Taking the nth root of the absolute value of the nth term, we have:

rn = [(n!) / (nn)](1/n)

We can simplify this expression by using Stirling’s formula:

n! ~ sqrt(2πn) (n/e)n

Thus, we have:

rn = [(n!) / (nn)](1/n)

= [sqrt(2πn) (n/e)n / nn](1/n)

= [(sqrt(2πn) / n) (1/e)n](1/n)

= [(sqrt(2πn) / n) / e]

Taking the limit as n approaches infinity, we have:

L = lim(n→∞) [(sqrt(2πn) / n) / e]

= lim(n→∞) [sqrt(2πn) / (n * e)]

= 0

Since L < 1, the series ∑ (n!) / (nn) converges absolutely.

In conclusion, Cauchy’s Root Test is a powerful tool for determining the convergence or divergence of infinite series. This test is particularly useful when the terms of the series have a complicated form or when other tests fail to provide conclusive results. By finding the nth root of the absolute value of the nth term, we can determine the value of L and determine whether the series converges or diverges.

# Describe D’Alembert’s Ratio Test

D’Alembert’s Ratio Test, also known as the Ratio Test or the Cauchy Ratio Test, is a useful tool for determining the convergence or divergence of infinite series. This test is particularly effective for series with positive terms.

To apply D’Alembert’s Ratio Test, we need to find the limit of the ratio of successive terms of the series:

L = lim(n→∞) |(a{n+1}) / an|

where an is the nth term of the series.

If L < 1, then the series converges absolutely. If L > 1, then the series diverges. If L = 1, then the test is inconclusive, and we need to apply other tests to determine convergence or divergence.

Example:

Consider the series ∑ (1/2)n. We can use D’Alembert’s Ratio Test to determine if the given series converges or diverges.

Taking the limit of the ratio of successive terms, we have:

L = lim(n→∞) |((1/2)(n+1)) / (1/2)n|

= lim(n→∞) (1/2)

= 1/2

Since L < 1, the series ∑ (1/2)n converges absolutely.

In conclusion, D’Alembert’s Ratio Test is a useful tool for determining the convergence or divergence of infinite series. By finding the limit of the ratio of successive terms, we can determine the value of L and determine whether the series converges or diverges. This test is particularly effective for series with positive terms.

# Examine the Convergence/Divergence of Infinite Series using D’Alembert’s Ratio Test

D’Alembert’s Ratio Test, also known as the Ratio Test or the Cauchy Ratio Test, is a useful tool for examining the convergence or divergence of infinite series. To use this test, we need to evaluate the limit of the ratio of successive terms of the series. If the limit exists and is less than 1, then the series converges absolutely. If the limit exists and is greater than 1, then the series diverges. If the limit is equal to 1, then the test is inconclusive, and we need to apply other tests to determine convergence or divergence.

Let’s take a look at some examples to better understand how to use D’Alembert’s Ratio Test.

Example 1:

Consider the series ∑ (n+1) / n!. We can use D’Alembert’s Ratio Test to determine if the given series converges or diverges.

Taking the limit of the ratio of successive terms, we have:

L = lim(n→∞) |((n+2) / (n+1)!) / ((n+1) / n!)|

= lim(n→∞) ((n+2) / (n+1)) * (n! / (n+1)!)

= lim(n→∞) (n+2) / (n+1)

= 1

Since L = 1, the Ratio Test is inconclusive. We need to apply other tests to determine the convergence or divergence of the series.

Example 2:

Consider the series ∑ (n3 + 3n) / 2n. We can use D’Alembert’s Ratio Test to determine if the given series converges or diverges.

Taking the limit of the ratio of successive terms, we have:

L = lim(n→∞) |(((n+1)3 + 3(n+1)) / 2(n+1)) / ((n3 + 3n) / 2n)|

= lim(n→∞) (n3 + 3n + 3n2 + 3n + 1) / (2n3)

= lim(n→∞) (1/2) * (1 + 3/n2 + 3/n3 + 1/n3)

= 1/2

Since L < 1, the series ∑ (n3 + 3n) / 2n converges absolutely.

In conclusion, D’Alembert’s Ratio Test is a powerful tool for examining the convergence or divergence of infinite series. By evaluating the limit of the ratio of successive terms, we can determine whether the series converges or diverges. If the limit is less than 1, the series converges absolutely, if the limit is greater than 1, the series diverges, and if the limit is equal to 1, then the test is inconclusive, and we need to apply other tests to determine convergence or divergence.

# Describe Raabe’s Test

Raabe’s test is a convergence test that is used to check the convergence of an infinite series. It is based on the comparison of the given series with a harmonic series. It is named after the German mathematician, August Ludwig von Raabe.

The Raabe’s test is applicable to infinite series of the form ∑an, where all the terms of the series are positive. The test provides a criterion for the convergence or divergence of a series based on the behavior of the ratio of consecutive terms of the series.

The test states that if the limit of the following expression exists and is positive, then the series converges:

R = lim(n→∞) (n[an/an+1]-1)

where an is the nth term of the series.

If the limit R is greater than 1, the series is divergent, and if R is less than 1, the series is convergent. If R is equal to 1, the test is inconclusive.

Example 1:

Consider the series ∑(n2)/(n3 + 1). To apply Raabe’s test, we need to calculate the ratio of consecutive terms:

Rn = n [(n2)/(n3 + 1)]/[(n+1)2/((n+1)3 + 1)] – 1

Taking the limit as n → ∞, we get:

R = limn→∞ (n[(n2)/(n3+1)]/[(n+1)2/((n+1)3+1)] – 1) = 2

Since R is greater than 1, the series diverges.

Example 2:

Consider the series ∑(n2)/(2n). To apply Raabe’s test, we need to calculate the ratio of consecutive terms:

Rn = n [(n2)/(2n)]/[(n+1)2/(2(n+1))] – 1

Taking the limit as n → ∞, we get:

R = limn→∞ (n[(n2)/(2n)]/[(n+1)2/(2(n+1))] – 1) = 1/2

Since R is less than 1, the series converges.

In conclusion, Raabe’s test is a useful convergence test for infinite series. It provides a criterion for the convergence or divergence of a series based on the behavior of the ratio of consecutive terms of the series. If the limit of the expression R exists and is positive, then the series converges. The test is named after the German mathematician, August Ludwig von Raabe.

# Examine the Convergence/Divergence of Infinite Series using Raabe’s Test

Raabe’s test is a convergence test used to examine the convergence or divergence of infinite series. It is based on comparing a given series with a harmonic series, which is a series of the form ∑1/n. The test is named after the German mathematician, August Ludwig von Raabe.

The Raabe’s test is applicable to infinite series of the form ∑an, where all the terms of the series are positive. The test provides a criterion for the convergence or divergence of a series based on the behavior of the ratio of consecutive terms of the series.

The test states that if the limit of the following expression exists and is positive, then the series converges:

R = limn→∞ (n[an/an+1]-1)

where an is the nth term of the series.

If the limit R is greater than 1, the series is divergent, and if R is less than 1, the series is convergent. If R is equal to 1, the test is inconclusive.

Example 1:

Consider the series ∑(n2)/(n3 + 1). To apply Raabe’s test, we need to calculate the ratio of consecutive terms:

Rn = n [(n2)/(n3 + 1)]/[(n+1)2/((n+1)3 + 1)] – 1

Taking the limit as n → ∞, we get:

R = limn→∞ (n[(n2)/(n3+1)]/[(n+1)2/((n+1)3+1)] – 1) = 2

Since R is greater than 1, the series diverges.

Example 2:

Consider the series ∑(n2)/(2n). To apply Raabe’s test, we need to calculate the ratio of consecutive terms:

Rn = n [(n2)/(2n)]/[(n+1)2/(2(n+1))] – 1

Taking the limit as n → ∞, we get:

R = limn→∞ (n[(n2)/(2n)]/[(n+1)2/(2(n+1))] – 1) = 1/2

Since R is less than 1, the series converges.

Example 3:

Consider the series ∑(n!)/(nn). To apply Raabe’s test, we need to calculate the ratio of consecutive terms:

Rn = n [(n!)/(nn)]/[(n+1)!/((n+1)(n+1))] – 1

Taking the limit as n → ∞, we get:

R = limn→∞ (n[(n!)/(nn)]/[(n+1)!/((n+1)(n+1))] – 1) = 1/e

Since R is less than 1, the series converges.

In conclusion, Raabe’s test is a useful convergence test for infinite series. It provides a criterion for the convergence or divergence of a series based on the behavior of the ratio of consecutive terms of the series. If the limit of the expression R exists and is positive, then the series converges. The test is named after the German mathematician, August Ludwig von Raabe.

# Describe Logarithmic Test

The logarithmic test is a convergence test used to examine the convergence or divergence of infinite series. It is a modification of the comparison test and is particularly useful for series in which the terms decrease very slowly.

The logarithmic test states that if we have a series of the form ∑an, where an is a positive sequence of terms, and if the sequence an is decreasing and tends to zero, then the series converges if and only if the series of the logarithms of the terms of the sequence converges. That is, if the series ∑log(an) converges, then the series ∑an also converges.

The proof of the logarithmic test is based on the properties of logarithms. Specifically, we can use the fact that log(n+1) – log(n) approaches zero as n approaches infinity to show that if the series of the logarithms converges, then the original series also converges.

Example 1:

Consider the series ∑(1/np), where p is a positive constant. We can use the logarithmic test to determine the convergence of this series. Taking the logarithm of the terms, we get log(an) = -p log(n). Therefore, the series of logarithms becomes ∑(-p log(n)). This series converges if and only if p > 1. Thus, the original series converges if and only if p > 1.

Example 2:

Consider the series ∑(1/log(n)). We can use the logarithmic test to determine the convergence of this series. Taking the logarithm of the terms, we get log(an) = log(1/log(n)) = -log(log(n)). Therefore, the series of logarithms becomes ∑(-log(log(n))). This series converges, and hence the original series also converges.

Example 3:

Consider the series ∑(1/n). We can use the logarithmic test to determine the convergence of this series. Taking the logarithm of the terms, we get log(an) = log(1/n) = -log(n). Therefore, the series of logarithms becomes ∑(-log(n)), which diverges. Hence, the original series also diverges.

In conclusion, the logarithmic test is a useful convergence test for infinite series. It is a modification of the comparison test and is particularly useful for series in which the terms decrease very slowly. The test states that the series of the logarithms of the terms of a sequence converges if and only if the original series also converges. The proof of the test is based on the properties of logarithms.

# Examine the Convergence/Divergence of Infinite Series using Logarithmic Test

The logarithmic test is a convergence test used to examine the convergence or divergence of infinite series. It is a modification of the comparison test and is particularly useful for series in which the terms decrease very slowly.

The logarithmic test states that if we have a series of the form ∑an, where an is a positive sequence of terms, and if the sequence an is decreasing and tends to zero, then the series converges if and only if the series of the logarithms of the terms of the sequence converges. That is, if the series ∑log(an) converges, then the series ∑an also converges.

To examine the convergence or divergence of an infinite series using the logarithmic test, we need to follow the following steps:

Step 1: Determine if the sequence of terms an is positive, decreasing and tends to zero as n goes to infinity.

Step 2: Take the logarithm of each term to obtain the sequence {log(an)}.

Step 3: Determine if the series of logarithms ∑log(an) converges or diverges.

Step 4: If the series of logarithms converges, then the original series ∑an converges. If the series of logarithms diverges, then the original series ∑an also diverges.

Example 1:

Consider the series ∑(1/n2). To examine the convergence of this series using the logarithmic test, we take the logarithm of the terms to get log(an) = -2log(n). Thus, the series of logarithms becomes ∑(-2log(n)). This series converges, since it is a constant times the series of the natural logarithms of the positive integers, which is a well-known convergent series. Therefore, the original series ∑(1/n2) also converges.

Example 2:

Consider the series ∑(1/n). To examine the convergence of this series using the logarithmic test, we take the logarithm of the terms to get log(an) = -log(n). Thus, the series of logarithms becomes ∑(-log(n)). This series diverges, since it is a constant times the harmonic series, which is a well-known divergent series. Therefore, the original series ∑(1/n) also diverges.

Example 3:

Consider the series ∑(1/(n(log(n))2)). To examine the convergence of this series using the logarithmic test, we take the logarithm of the terms to get log(an) = -2log(n) – 2log(log(n)). Thus, the series of logarithms becomes ∑(-2log(n) – 2log(log(n))). This series converges, since it is a constant times the series of the natural logarithms of the positive integers, which is a well-known convergent series. Therefore, the original series ∑(1/(n(log(n))2)) also converges.

In conclusion, the logarithmic test is a useful convergence test for infinite series. It is a modification of the comparison test and is particularly useful for series in which the terms decrease very slowly. To examine the convergence or divergence of a series using the logarithmic test, we need to determine if the series of logarithms converges or diverges.

# Describe Cauchy’s Integral Test

Cauchy’s integral test is a convergence test used to determine the convergence or divergence of infinite series. It is based on the comparison of the sum of a series to an improper integral.

The integral test states that if we have a series of the form ∑an, where an is a positive sequence of terms, and if f(x) is a continuous, positive, decreasing function on [1, ∞) such that f(n) = an for all n ≥ 1, then the series ∑an converges if and only if the improper integral f(x)dx converges.

In other words, the test states that the convergence or divergence of an infinite series can be determined by comparing it to the convergence or divergence of an improper integral.

To apply the Cauchy’s integral test, we need to follow the following steps:

Step 1: Find a positive, continuous, and decreasing function f(x) such that f(n) = an for all n ≥ 1.

Step 2: Determine if the improper integral f(x)dx converges or diverges.

Step 3: If the improper integral converges, then the original series ∑an also converges. If the improper integral diverges, then the original series ∑an also diverges.

Example 1:

Consider the series ∑(1/n2). To apply Cauchy’s integral test to this series, we let f(x) = 1/x2. Then, f(n) = 1/n2 = an for all n ≥ 1. The improper integral (1/x2)dx = [(-1/x)|1] = 1, which converges. Therefore, the original series ∑(1/n2) also converges.

Example 2:

Consider the series ∑(1/n). To apply Cauchy’s integral test to this series, we let f(x) = 1/x. Then, f(n) = 1/n = an for all n ≥ 1. The improper integral (1/x)dx = [ln(x)|1] = ∞, which diverges. Therefore, the original series ∑(1/n) also diverges.

Example 3:

Consider the series ∑(1/(n2 + 1)). To apply Cauchy’s integral test to this series, we let f(x) = 1/(x2 + 1). Then, f(n) = 1/(n2 + 1) = an for all n ≥ 1. The improper integral (1/(x2 + 1))dx = [(arctan(x))|1 ∞] = (π/2) – (π/4) = π/4, which converges. Therefore, the original series ∑(1/(n2 + 1)) also converges.

In conclusion, Cauchy’s integral test is a useful convergence test for infinite series. It allows us to compare the sum of a series to an improper integral, and thereby determine its convergence or divergence. The integral test can be especially useful for determining the convergence of series with terms that are difficult to work with using other methods.

# Examine the Convergence/Divergence of Infinite Series using Cauchy’s Integral Test

Cauchy’s Integral Test is a convergence test used to examine the convergence or divergence of infinite series. It is based on the comparison of the sum of a series to an improper integral. The test states that if we have a series of the form ∑an, where an is a positive sequence of terms, and if f(x) is a continuous, positive, decreasing function on [1, ∞) such that f(n) = an for all n ≥ 1, then the series ∑an converges if and only if the improper integral f(x)dx converges.

To apply Cauchy’s Integral Test, we follow the following steps:

Step 1: Find a positive, continuous, and decreasing function f(x) such that f(n) = an for all n ≥ 1.

Step 2: Determine if the improper integralf(x)dx converges or diverges.

Step 3: If the improper integral converges, then the original series ∑an also converges. If the improper integral diverges, then the original series ∑an also diverges.

Example 1:

Determine the convergence or divergence of the series ∑(1/n2). To apply Cauchy’s Integral Test, we let f(x) = 1/x2. Then, f(n) = 1/n2 = an for all n ≥ 1. The improper integral(1/x2)dx = [(-1/x)|1] = 1, which converges. Therefore, the original series ∑(1/n2) also converges.

Example 2:

Determine the convergence or divergence of the series ∑(1/n). To apply Cauchy’s Integral Test, we let f(x) = 1/x. Then, f(n) = 1/n = an for all n ≥ The improper integral (1/x)dx = [ln(x)|1] = ∞, which diverges. Therefore, the original series ∑(1/n) also diverges.

Example 3:

Determine the convergence or divergence of the series ∑(1/(n2 + 1)). To apply Cauchy’s Integral Test, we let f(x) = 1/(x2 + 1). Then, f(n) = 1/(n2 + 1) = an for all n ≥ 1. The improper integral (1/(x2 + 1))dx = [(arctan(x))|1] = (π/2) – (π/4) = π/4, which converges. Therefore, the original series ∑(1/(n2 + 1)) also converges.

In conclusion, Cauchy’s Integral Test is a useful convergence test for infinite series. It allows us to compare the sum of a series to an improper integral, and thereby determine its convergence or divergence. The integral test can be especially useful for determining the convergence of series with terms that are difficult to work with using other methods. However, it may not always be applicable or easy to use in practice. Therefore, it is important to understand other convergence tests as well.

# Describe Alternating Series

An alternating series is a series whose terms alternate in sign, meaning that the signs of the terms alternate between positive and negative. An alternating series takes the form ∑(-1)n bn, where bn is a positive sequence of terms. The alternating series test is a convergence test that can be used to determine whether an alternating series converges or diverges.

The alternating series test states that if a series satisfies the following conditions, then it converges:

1. The terms of the series are positive and decreasing: b1 ≥ b2 ≥ b3 ≥ …
2. The terms of the series approach zero: limn→∞ bn = 0.
3. The terms of the series alternate in sign: the signs of the terms of the series alternate between positive and negative.

To apply the alternating series test, we follow the following steps:

Step 1: Check if the series satisfies the conditions of the alternating series test.

Step 2: Determine if the limit of the terms of the series approaches zero.

Step 3: If the series satisfies the conditions of the alternating series test and the limit of the terms of the series approaches zero, then the series converges.

Example 1:

Determine whether the series ∑(-1)n (1/n) converges or diverges. This series satisfies the conditions of the alternating series test since the terms of the series are positive and decreasing, the limit of the terms of the series approaches zero, and the terms of the series alternate in sign. Therefore, by the alternating series test, the series converges.

Example 2:

Determine whether the series ∑(-1)n (1/(2n + 1)) converges or diverges. This series satisfies the conditions of the alternating series test since the terms of the series are positive and decreasing, the limit of the terms of the series approaches zero, and the terms of the series alternate in sign. Therefore, by the alternating series test, the series converges.

Example 3:

Determine whether the series ∑(-1)n (n/(n + 1)) converges or diverges. This series does not satisfy the conditions of the alternating series test since the terms of the series are not decreasing. Therefore, we cannot use the alternating series test to determine whether the series converges or diverges.

In conclusion, an alternating series is a series whose terms alternate in sign, and the alternating series test is a convergence test that can be used to determine whether an alternating series converges or diverges. The alternating series test can be a useful tool for determining the convergence of certain series, but it is important to understand that not all series can be tested using this method. It is important to be familiar with other convergence tests as well.

# Explain Leibniz’s Rule

Leibnitz’s rule, also known as the alternating series remainder theorem, provides a method for estimating the remainder of an alternating series. An alternating series is a series whose terms alternate in sign, meaning that the signs of the terms alternate between positive and negative. Leibnitz’s rule can be used to determine how well an alternating series converges.

Leibnitz’s rule states that the remainder of an alternating series, R n, is bounded by the magnitude of the (n+1)th term of the series. In other words, if we denote the nth term of an alternating series as an and the (n+1)th term as an+1, then:

|Rn| ≤ |an+1|

This means that the remainder of the series is less than or equal to the absolute value of the (n+1)th term of the series. This can be a useful tool in estimating the accuracy of the sum of an alternating series.

To use Leibnitz’s rule to estimate the sum of an alternating series, we first need to find the sum of a portion of the series. Then, we can use Leibnitz’s rule to estimate the remainder of the series and add it to the partial sum to get an estimate of the total sum.

Example:

Determine how many terms are required to find the sum of the series ∑(-1)n (1/n2) to within an error of 0.001.

We know that the nth term of the series is an = (-1)n (1/n2), which is positive and decreasing. To find the number of terms required to find the sum of the series within an error of 0.001, we need to use Leibnitz’s rule to estimate the remainder of the series.

Since the terms of the series are decreasing, we know that the remainder of the series is less than the magnitude of the (n+1)th term of the series. Therefore, we can set |Rn| ≤ |an+1| and solve for n as follows:

|(-1)(n+1) (1/(n+1)2)| ≤ 0.001

1/(n+1)2 ≤ 0.001

n+1 ≥ 31.623

n ≥ 30.623

Therefore, we need at least 31 terms to find the sum of the series within an error of 0.001.

In conclusion, Leibnitz’s rule is a useful tool for estimating the remainder of an alternating series. By using Leibnitz’s rule, we can determine how well an alternating series converges and estimate the accuracy of the sum of the series. Leibnitz’s rule can be used in conjunction with other convergence tests to provide a more accurate estimate of the sum of a series.

# Describe Absolute and Conditional Convergence

When discussing the convergence of a series, it is important to consider whether the series converges absolutely or conditionally. A series is said to converge absolutely if the sum of the absolute values of its terms converges. In contrast, a series is said to converge conditionally if it converges, but its absolute values do not converge.

Mathematically, a series ∑an converges absolutely if ∑|an| converges. A series ∑an converges conditionally if ∑an converges, but ∑|an | diverges.

Absolute convergence is a stronger form of convergence than conditional convergence. If a series converges absolutely, then it is guaranteed to converge. However, if a series converges conditionally, it may or may not converge depending on the order in which its terms are added.

One important property of absolute convergence is that it is invariant under rearrangement. In other words, if a series converges absolutely, then its sum is the same regardless of the order in which its terms are added. This is not true for conditionally convergent series. If a conditionally convergent series is rearranged, the sum of the series may change.

Examples:

1. The series ∑(-1)n (1/n) is a well-known example of a conditionally convergent series. The absolute value of each term is 1/n, and the series ∑(1/n) is known to diverge. However, the series converges because the terms alternate in sign, causing the positive and negative terms to cancel each other out. In this case, the series converges conditionally but not absolutely.
2. The series ∑(1/n2) is an example of a series that converges absolutely. The absolute value of each term is 1/n2, and the series ∑(1/n2) is known to converge. Therefore, the series converges absolutely.

In conclusion, absolute convergence and conditional convergence are important concepts in the study of series. A series that converges absolutely is guaranteed to converge, and its sum is invariant under rearrangement. In contrast, a series that converges conditionally may or may not converge depending on the order in which its terms are added. Understanding these concepts is important for analyzing the convergence of series and their applications in mathematics and other fields.