Question:medium

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

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$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \log 2$.

Updated On: Apr 8, 2026

$e^{-1}$
$\log 2 - 1$
$1$
$0$

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The Correct Option is B

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

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The Correct Option is B

Solution and Explanation

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