Σ_(n=1) ^∞ (((−1)^(n+1) )/(n(2n+1)))=?


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Question Number 161622 by amin96 last updated on 20/Dec/21

$\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n (2 n + 1)} = ?$
	Answered by mathmax by abdo last updated on 20/Dec/21
	$(S/2)=−Σ_(n=1) ^∞ (((−1)^n )/(2n(2n+1)))=−Σ_(n=1) ^∞ (−1)^n {(1/(2n))−(1/(2n+1))} =−(1/2)Σ_(n=1) ^∞ (((−1)^n )/n)−Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) =((log2)/2)−((π/4)−1)=((log2)/2)−(π/4)+1$
	$\frac{S}{2} = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n (2 n + 1)} = - \sum_{n = 1}^{\infty} {(- 1)}^{n} {\frac{1}{2 n} - \frac{1}{2 n + 1}}$ $= - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}$ $= \frac{\log 2}{2} - (\frac{π}{4} - 1) = \frac{\log 2}{2} - \frac{π}{4} + 1$
	Commented by mathmax by abdo last updated on 20/Dec/21

	$\Rightarrow S = \log 2 - \frac{π}{2} + 2$
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