calculate Σ_(n=1) ^∞ (((−1)^n )/(4n^2 −1)) .


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Question Number 43260 by maxmathsup by imad last updated on 08/Sep/18

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n^{2} - 1} .$
Commented by maxmathsup by imad last updated on 11/Sep/18
$let S =Σ_(n=1) ^∞ (((−1)^n )/(4n^2 −1)) S =(1/2) Σ_(n=1) ^∞ (−1)^n { (1/(2n−1))−(1/(2n+1))} =(1/2)Σ_(n=1) ^∞ (((−1)^n )/(2n−1)) −(1/2) Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) but Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) =(π/4) −1 Σ_(n=1) ^∞ (((−1)^n )/(2n−1)) =_(n=p+1) Σ_(p=0) ^∞ (((−1)^(p+1) )/(2p+1)) =−(π/4) ⇒ S = (1/2){−(π/4)}−(1/2){(π/4)−1} =−(π/8) −(π/8) +(1/2) = (1/2) −(π/4) S=(1/2) −(π/4) .$
$l e t S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{4 n^{2} - 1}$ $S = \frac{1}{2} \sum_{n = 1}^{\infty} {(- 1)}^{n} {\frac{1}{2 n - 1} - \frac{1}{2 n + 1}} = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n - 1} - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}$ $b u t \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4} - 1$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n - 1} =_{n = p + 1} \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p + 1}}{2 p + 1} = - \frac{π}{4} \Rightarrow$ $S = \frac{1}{2} {- \frac{π}{4}} - \frac{1}{2} {\frac{π}{4} - 1} = - \frac{π}{8} - \frac{π}{8} + \frac{1}{2} = \frac{1}{2} - \frac{π}{4}$ $S = \frac{1}{2} - \frac{π}{4} .$
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