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Chapter 7: Problem 45

Find the sum of the convergent series by using a well-known function. Identify the function and explain how you obtained the sum. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2^{n} n} $$

Short Answer

Expert verified
The sum of the given series is \( \frac{3}{2}\log 2\).

Step by step solution

01

Identify the Function

The given series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2^{n} n}\) can be recognized as the derivative of the geometric series sum formula \(\frac{1}{1-x}\), where \(x\) is equal to \(-\frac{1}{2}\). Therefore, we begin by differentiating this function to see if we get some form of our given series.
02

Differentiate the Function

The derivative of the function \(f(x) = \frac{1}{1-x}\) with respect to \(x\) is \(f'(x) = \frac{1}{(1-x)^2}\). Substitution \(x=-\frac{1}{2}\) into \(f'(x)\) leads to -4, which does not directly match our given series.
03

Modify the Derivative Rule and Integrate

Upon examining the pattern of our given series, we realize that for general term \((-1)^{n+1} \frac{1}{2^{n} n}\), we need to modify our derivative rule as follows: The derivative of \(-\log(1-x)\) is \(\frac{1}{1-x}\). Therefore, by differentiating our function \(f(x)\) as \(f(x) = -\log(1-x)\) and substituting \(x=-\frac{1}{2}\), a pattern equivalent to our series is obtained, resulting in \(\sum_{n=1}^{\infty}\frac{x^n}{n}\). Using this, we can integrate from \(0\) to \(-\frac{1}{2}\) to find the sum of the series.
04

Calculate the Sum

Integrating \(-\log(1-x)\) from \(0\) to \(-\frac{1}{2}\) gives us the sum of the series. The definite integral is \(\int_{0}^{-1/2} -\log(1-x) dx\), which equals to \(\log 2 - (-\frac{1}{2}\log 2)\), simplifying to \( \frac{3}{2}\log 2\) which is the sum of the original series.

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Most popular questions from this chapter

Find the sum of the convergent series. $$ \sum_{n=1}^{\infty} \frac{1}{(2 n+1)(2 n+3)} $$

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(1+\frac{k}{n}\right)^{n} $$

Describe the difference between \(\lim _{n \rightarrow \infty} a_{n}=5\) and \(\sum_{n=1}^{\infty} a_{n}=5\).

Give an example of a sequence satisfying the condition or explain why no such sequence exists. (Examples are not unique.) A sequence that converges to \(\frac{3}{4}\)

(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{81} $$
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