let S_n = Σ_(k=2) ^n (((−1)^k )/(k^2 −1)) 1) calculate S_n interms of H_n ( H_n =Σ_(k=1) ^n (1/k)) 2) find lim_(n→+∞) S


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Question Number 40046 by abdo mathsup 649 cc last updated on 15/Jul/18

$l e t S_{n} = \sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k^{2} - 1}$ $1) c a l c u l a t e S_{n} i n t e r m s o f H_{n}$ $(H_{n} = \sum_{k = 1}^{n} \frac{1}{k})$ $2) f i n d {l i m}_{n \to + \infty} S_{n}$
Commented by math khazana by abdo last updated on 26/Jul/18
$1) we have S_n = Σ_(k=2) ^n (((−1)^k )/(k^2 −1)) =(1/2)Σ_(k=2) ^n (−1)^k { (1/(k−1)) −(1/(k+1))} =(1/2){ Σ_(k=2) ^n (((−1)^k )/(k−1)) −Σ_(k=2) ^n (((−1)^k )/(k+1))} but Σ_(k=2) ^n (((−1)^k )/(k−1)) =Σ_(k=1) ^(n−1) (((−1)^(k+1) )/k) =Σ_(p=1) ^([((n−1)/2)]) ((−1)/(2p)) + Σ_(p=0) ^([((n−2)/2)]) (1/(2p+1)) =−(1/2)H_([((n−1)/2)]) +Σ_(p=0) ^([((n−2)/2)]) (1/(2p+1)) Σ_(k=2) ^n (((−1)^k )/(k+1)) = Σ_(k=3) ^(n+1) (((−1)^(k−1) )/k) =Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) −(1−(1/2))=Σ_(k=1) ^(n+1) (((−1)^(k−1) )/k) −(1/2) =Σ_(p=1) ^([((n+1)/2)]) ((−1)/(2p)) +Σ_(p=0) ^([(n/2)]) (1/(2p+1)) −(1/2) ⇒ S_n = (1/2){−(1/2)H_([((n−1)/2)]) +Σ_(p=0) ^([((n−2)/2)]) (1/(2p+1)) +(1/2)H_([((n+1)/2)]) −Σ_(p=0) ^([(n/2)]) (1/(2p+1)) +(1/2)}$
$1) w e h a v e S_{n} = \sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k^{2} - 1}$ $= \frac{1}{2} \sum_{k = 2}^{n} {(- 1)}^{k} {\frac{1}{k - 1} - \frac{1}{k + 1}}$ $= \frac{1}{2} {\sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k - 1} - \sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k + 1}} b u t$ $\sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k - 1} = \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k + 1}}{k}$ $= \sum_{p = 1}^{[\frac{n - 1}{2}]} \frac{- 1}{2 p} + \sum_{p = 0}^{[\frac{n - 2}{2}]} \frac{1}{2 p + 1}$ $= - \frac{1}{2} H_{[\frac{n - 1}{2}]} + \sum_{p = 0}^{[\frac{n - 2}{2}]} \frac{1}{2 p + 1}$ $\sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k + 1} = \sum_{k = 3}^{n + 1} \frac{{(- 1)}^{k - 1}}{k}$ $= \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} - (1 - \frac{1}{2}) = \sum_{k = 1}^{n + 1} \frac{{(- 1)}^{k - 1}}{k} - \frac{1}{2}$ $= \sum_{p = 1}^{[\frac{n + 1}{2}]} \frac{- 1}{2 p} + \sum_{p = 0}^{[\frac{n}{2}]} \frac{1}{2 p + 1} - \frac{1}{2} \Rightarrow$ $S_{n} = \frac{1}{2} {- \frac{1}{2} H_{[\frac{n - 1}{2}]} + \sum_{p = 0}^{[\frac{n - 2}{2}]} \frac{1}{2 p + 1} + \frac{1}{2} H_{[\frac{n + 1}{2}]}$ $- \sum_{p = 0}^{[\frac{n}{2}]} \frac{1}{2 p + 1} + \frac{1}{2}}$
Commented by math khazana by abdo last updated on 26/Jul/18
$2) we have S_n =(1/2){ Σ_(k=2) ^n (((−1)^k )/(k−1)) −Σ_(k=2) ^n (((−1)^k )/(k+1))} but Σ_(k=2) ^n (((−1)^k )/(k−1)) =Σ_(k=1) ^(n−1) (((−1)^(k+1) )/k) =−Σ_(k=1) ^(n−1) (((−1)^k )/k) Σ_(k=2) ^n (((−1)^k )/(k+1)) = Σ_(k=3) ^(n+1) (((−1)^(k−1) )/k) =−Σ_(k=3) ^(n+1) (((−1)^k )/k) =−{ Σ_(k=1) ^(n−1) (((−1)^k )/k) +(((−1)^n )/n) +(((−1)^(n+1) )/(n+1)) −(((−1))/1) −(1/2)} = −Σ_(k=1) ^(n−1) (((−1)^k )/k) −(((−1)^n )/n) +(((−1)^n )/(n+1)) −(1/2) ⇒ S_n =(1/2){−Σ_(k=1) ^(n−1) (((−1)^k )/k) +Σ_(k=1) ^(n−1) (((−1)^k )/k) +(((−1)^n )/n) −(((−1)^n )/(n+1)) +(1/2)} S_n = (1/2){ (((−1)^n )/n) −(((−1)^n )/(n+1)) +(1/2)} ⇒ lim_(n→+∞) S_n =(1/4) .$
$2) w e h a v e S_{n} = \frac{1}{2} {\sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k - 1} - \sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k + 1}}$ $b u t \sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k - 1} = \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k + 1}}{k} = - \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{k}$ $\sum_{k = 2}^{n} \frac{{(- 1)}^{k}}{k + 1} = \sum_{k = 3}^{n + 1} \frac{{(- 1)}^{k - 1}}{k}$ $= - \sum_{k = 3}^{n + 1} \frac{{(- 1)}^{k}}{k} = - {\sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{k} + \frac{{(- 1)}^{n}}{n} + \frac{{(- 1)}^{n + 1}}{n + 1}$ $- \frac{(- 1)}{1} - \frac{1}{2}}$ $= - \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{k} - \frac{{(- 1)}^{n}}{n} + \frac{{(- 1)}^{n}}{n + 1} - \frac{1}{2} \Rightarrow$ $S_{n} = \frac{1}{2} {- \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{k} + \sum_{k = 1}^{n - 1} \frac{{(- 1)}^{k}}{k} + \frac{{(- 1)}^{n}}{n}$ $- \frac{{(- 1)}^{n}}{n + 1} + \frac{1}{2}}$ $S_{n} = \frac{1}{2} {\frac{{(- 1)}^{n}}{n} - \frac{{(- 1)}^{n}}{n + 1} + \frac{1}{2}} \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = \frac{1}{4} .$
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