Problem 45 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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.