Σ_(n=1) ^∞ (((−1)^(n+1) )/(n(n+2)))=?


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Question Number 161623 by amin96 last updated on 20/Dec/21

$\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n (n + 2)} = ?$
	Answered by TheSupreme last updated on 20/Dec/21

	$\frac{A}{n} + \frac{B}{n + 2} = \frac{{(- 1)}^{n + 1}}{n (n + 2)}$ $A n + 2 A + B n = {(- 1)}^{n + 1}$ $A = - B = \frac{{(- 1)}^{n + 1}}{2}$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{2} \frac{1}{n} - \frac{{(- 1)}^{n + 1}}{2} \frac{1}{n + 2}$ $s_{n} = \frac{1}{2} - \frac{1}{3} + \frac{{(- 1)}^{n}}{n + 2}$ $l i m s_{n} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$
	Commented by mr W last updated on 21/Dec/21

	$h o w d i d y o u c o m e f r o m$ $A n + 2 A + B n = {(- 1)}^{n + 1}$ $t o$ $A = - B = \frac{{(- 1)}^{n + 1}}{2} ?$
	Answered by mathmax by abdo last updated on 20/Dec/21
	$S_n =Σ_(k=1) ^n (((−1)^(k+1) )/(k(k+2))) ⇒S_n =(1/2)Σ_(k=1) ^n ((1/k)−(1/(k+2)))(−1)^(k+1) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k) +(1/2)Σ_(k=1) ^(n ) (((−1)^k )/(k+2)) (→k+2=p) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k)+(1/2)Σ_(p=3) ^(n+2) (((−1)^p )/p) =−(1/2)Σ_(k=1) ^n (((−1)^k )/k)+(1/2){Σ_(p=1) ^n (((−1)^p )/p)−(−1+(1/2))+(((−1)^n )/(n+2))} =(1/2){(1/2)+(((−1)^n )/(n+2))} ⇒lim_(n→+∞) S_n =(1/4)$
	$S_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k + 1}}{k (k + 2)} \Rightarrow S_{n} = \frac{1}{2} \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 2}) {(- 1)}^{k + 1}$ $= - \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} + \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k + 2} (\to k + 2 = p)$ $= - \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} + \frac{1}{2} \sum_{p = 3}^{n + 2} \frac{{(- 1)}^{p}}{p}$ $= - \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} + \frac{1}{2} {\sum_{p = 1}^{n} \frac{{(- 1)}^{p}}{p} - (- 1 + \frac{1}{2}) + \frac{{(- 1)}^{n}}{n + 2}}$ $= \frac{1}{2} {\frac{1}{2} + \frac{{(- 1)}^{n}}{n + 2}} \Rightarrow \lim_{n \to + \infty} S_{n} = \frac{1}{4}$
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