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Absolute Convergence

Here, we can learn about series and understand the concept of absolutely convergent. Since we know that a series is the sum of the terms of a sequence of numbers ${a_1, a_2, a_3, a_4, a_5......}$ the $n^{th}$ partial sum $S_n$ is the sum of the first n terms of the sequence, that is $S_{n} = \sum_{n = 0}^{\infty} a_{n}$

A series is convergent if the sequence of its partial sums ${S_1, S_2, S_3, S_4,.......}$ converges. A series that is not convergent is said to be divergent. The absolutely convergence is the convergence test of the alternating series. An alternating series is the sum of the terms of a sequence numbers with alternating positive and negative signs, which is denoted by $\sum_{n = 1}^{\infty} -1^{n - 1} a_{n}$

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Absolute Convergence Test

To know about the test of the absolute convergence, first we have to know about the definition of absolute convergence.

A series $\sum_{n = 0}^{\infty} a_{n}$ is said to converge absolutely if the series of absolute values $\sum_{n = 0}^{\infty} | a_{n} |$ converges. For any sequence ${a_1, a_2, a_3, a_4, a_5,......}$, $a_{n} \leq \left | a_{n} \right |$ for all n.

Therefore, $\sum_{n = 0}^{\infty} a_{n} \leq \sum_{n = 0}^{\infty} \left | a_{n} \right |$

Suppose, we have the series $\sum_{n = 1}^{\infty} a_{n}$

Define, $L = \lim_{n \to \infty} \left | \frac{a_{n + 1}}{a_n} \right |$

Then,

• If L < 1, the series is absolutely convergent (and hence convergent).
• If L > 1, the series is divergent.
• If L = 1, the series may be divergent, conditionally convergent, or absolutely convergent.
1. The product of two absolute convergent series is absolutely convergent series.
2. Every absolutely convergent series is convergent but the convergent series may or may not converge absolutely.

Absolute and Conditional Convergence

We know that a series $\sum x_{n}$ is said to be absolute convergent, if $\sum \left | x_{n} \right |$ is convergent.

For example, the series $\sum x_{n} = 1 - \frac{1}{2} + \frac{1}{(2)^{2}} - \frac{1}{(2)^{3}} + .....$ is absolutely convergent, since $\sum x_{n} = 1 + \frac{1}{2} + \frac{1}{(2)^{2}} + \frac{1}{(2)^{3}} + .....$, being a geometric series with common ratio $\frac{1}{2}$ < 1, is convergent.

Conditional Convergence:
A series $\sum x_{n}$ is said to be conditionally convergent if
1. $\sum x_{n}$ is convergent
2. $\sum x_{n}$ is not absolutely convergent.

For example, the series $\sum x_{n} = 1 - \frac{1}{2} + \frac{1}{(3)} - \frac{1}{(4)} +.....$ is conditionally convergent since the given series is convergent and $I\sum x_{n}I = 1 + \frac{1}{2} + \frac{1}{(3)} + \frac{1}{(4)} + .....$ is not convergent, i.e. $\sum x_{n}$ is not absolutely convergent.

Every absolutely convergent series is convergent but the converse is not true.
Given the series $\sum_{n = 1}^{\infty }x_{n}$
• If $\sum_{n = 1}^{\infty }x_{n}$ is absolutely convergent and its value is s then any rearrangement of $\sum_{n = 1}^{\infty }x_{n}$ will also have a value of s.
• If $\sum_{n = 1}^{\infty }x_{n}$ is conditionally convergent and r is any real number then there is a rearrangement of $\sum_{n = 1}^{\infty }x_{n}$ whose value will be r.

Absolute Convergence Examples

Given below are some of the examples on absolute convergence.

Solved Examples

Question 1: Determine if the following series is convergent or divergent. Solution:
Here, $a_{n} = \frac{n^2}{(2n - 1)!}$ and $a_{n + 1} = \frac{(n + 1)^2}{(2(n + 1) - 1)!}$

By Ratio Test, find the limit L,  So, L < 1, by ratio test the series converge absolutely and hence converge.

Question 2: Show that the series $\sum_{n = 1}^{\infty} \frac{-1^{n + 1}}{n^2}$ is absolutely convergent.
Solution:
Since $\sum_{n = 1}^{\infty} | \frac{-1^{n + 1}}{n^2} | = \sum_{n = 1}^{\infty} \frac{1}{n^2}$

and the series $\sum_{n = 1}^{\infty} \frac{1}{n^2}$ is convergent.

Hence, the given series is absolutely convergent.

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