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


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Question Number 34688 by math khazana by abdo last updated on 09/May/18

$c s l c u l a t e \sum_{n = 2}^{\infty} l n (1 + \frac{{(- 1)}^{n}}{n})$
Commented by abdo mathsup 649 cc last updated on 14/May/18
$let put S_n = Σ_(k=2) ^n ln(1+(((−1)^k )/k)) S_n = ln{ Π_(k=2) ^n (1+(((−1)^k )/k))}=ln(w_n ) but w_n = Π_(k=2) ^n (1+(((−1)^k )/k)) = Π_(p=1) ^([(n/2)]) ( 1+ (1/(2p))) Π_(p=1) ^([((n−1)/2)]) (1−(1/(2p+1))) w_(2n) =Π_(p=1) ^n ((2p+1)/(2p)) Π_(p=1) ^(n−1) ((2p)/(2p+1)) = ((2n−1)/(2n−2)) Π_(p=1) ^(n−1) (1) = ((2n−1)/(2n−2)) →1(n→+∞) lim_(n→+∞) S_(2n) = 0 w_(2n+1) = Π_(p=1) ^n (1−(1/(2p+1)))Π_(p=1) ^n (1 +(1/(2p))) = Π_(p=1) ^n ((2p)/(2p+1)) Π_(p=1) ^n ((2p+1)/(2p)) =1 ⇒lim S_(2n+1) =0 so S_n →0(n→+∞)$
$l e t p u t S_{n} = \sum_{k = 2}^{n} l n (1 + \frac{{(- 1)}^{k}}{k})$ $S_{n} = l n {\prod_{k = 2}^{n} (1 + \frac{{(- 1)}^{k}}{k})} = l n (w_{n}) b u t$ $w_{n} = \prod_{k = 2}^{n} (1 + \frac{{(- 1)}^{k}}{k}) = \prod_{p = 1}^{[\frac{n}{2}]} (1 + \frac{1}{2 p}) \prod_{p = 1}^{[\frac{n - 1}{2}]} (1 - \frac{1}{2 p + 1})$ $w_{2 n} = \prod_{p = 1}^{n} \frac{2 p + 1}{2 p} \prod_{p = 1}^{n - 1} \frac{2 p}{2 p + 1}$ $= \frac{2 n - 1}{2 n - 2} \prod_{p = 1}^{n - 1} (1) = \frac{2 n - 1}{2 n - 2} \to 1 (n \to + \infty)$ ${l i m}_{n \to + \infty} S_{2 n} = 0$ $w_{2 n + 1} = \prod_{p = 1}^{n} (1 - \frac{1}{2 p + 1}) \prod_{p = 1}^{n} (1 + \frac{1}{2 p})$ $= \prod_{p = 1}^{n} \frac{2 p}{2 p + 1} \prod_{p = 1}^{n} \frac{2 p + 1}{2 p} = 1 \Rightarrow l i m S_{2 n + 1} = 0 s o$ $S_{n} \to 0 (n \to + \infty)$
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