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


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Question Number 43259 by maxmathsup by imad last updated on 08/Sep/18

$c a l c u l a t e \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - 1}$
Commented by maxmathsup by imad last updated on 09/Sep/18
$let S = Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1)) ⇒S=(1/2) Σ_(n=2) ^∞ (−1)^n {(1/(n−1)) −(1/(n+1))} ⇒ 2S = Σ_(n=2) ^∞ (((−1)^n )/(n−1)) −Σ_(n=2) ^∞ (((−1)^n )/(n+1)) but Σ_(n=2) ^∞ (((−1)^n )/(n−1)) = Σ_(n=1) ^∞ (((−1)^(n+1) )/n) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(2) (Σ_(n=1) ^∞ (x^n /n) = −ln(1−x) with −1<x<1) also Σ_(n=2) ^∞ (((−1)^n )/(n+1)) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −{1 −(1/2)} =ln(2) −(1/2) ⇒ 2S = ln(2) −ln(2)+(1/2) ⇒ S =(1/4) .$
$l e t S = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - 1} \Rightarrow S = \frac{1}{2} \sum_{n = 2}^{\infty} {(- 1)}^{n} {\frac{1}{n - 1} - \frac{1}{n + 1}}$ $\Rightarrow 2 S = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n - 1} - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n + 1} b u t$ $\sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n - 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2)$ $(\sum_{n = 1}^{\infty} \frac{x^{n}}{n} = - l n (1 - x) w i t h - 1 < x < 1) a l s o$ $\sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} - {1 - \frac{1}{2}}$ $= l n (2) - \frac{1}{2} \Rightarrow 2 S = l n (2) - l n (2) + \frac{1}{2} \Rightarrow S = \frac{1}{4} .$
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