Question:medium

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

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Memorize expansion of \(\log(1+x)\) for alternating series.
Updated On: Apr 18, 2026
  • \(e^{-1}\)
  • \(\log 2 - 1\)
  • 1
  • 0
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This problem involves the expansion of the logarithmic series \( \log(1+x) \).
: Key Formula or Approach:
The standard series expansion is: \[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \] For \( x = 1 \): \[ \log 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \] Step 2: Detailed Explanation:
The given summation is: \[ S = \sum_{n=1}^{\infty} \frac{(-1)^n}{n+1} \] Expand the sum: \[ S = \frac{-1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \] Compare this with the expansion for \( \log 2 \): \[ \log 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \] Notice that: \[ \log 2 = 1 + \left( -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right) \] \[ \log 2 = 1 + S \] Rearranging for \( S \): \[ S = \log 2 - 1 \].
Step 3: Final Answer:
The value of the sum is \( \log 2 - 1 \).
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