Question

\( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n+1} \) is equal to

• A. \(e^{-1}\)
• B. \(\log 2 - 1\)
• C. 1
• D. 0

Hint:

Memorize expansion of \(\log(1+x)\) for alternating series.

Answer 1

The Correct Option is B

Concept: \[ \log(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} \]

Step 1: Use expansion at \(x=1\).
\[ \log 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \]

Step 2: Shift index.
\[ \sum \frac{(-1)^{n-1}}{n+1} = \log 2 - 1 \]