let S_n = Σ_(k=1) ^n (((−1)^k )/k) 1) calculate S_n interms of H_n 2) find lim_(n→+∞) S_n 3) let W_n = Σ_(1≤i<j≤n) (((−1)^(i+j) )/(i.j)) prove that (W_n ) is convergent and calculste its limit.


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Question Number 40067 by abdo mathsup 649 cc last updated on 15/Jul/18

$l e t S_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k}$ $1) c a l c u l a t e S_{n} i n t e r m s o f H_{n}$ $2) f i n d {l i m}_{n \to + \infty} S_{n}$ $3) l e t W_{n} = \sum_{1 ⩽ i < j ⩽ n} \frac{{(- 1)}^{i + j}}{i . j}$ $p r o v e t h a t (W_{n}) i s c o n v e r g e n t a n d c a l c u l s t e i t s$ $l i m i t .$
Commented byabdo mathsup 649 cc last updated on 17/Jul/18
$S_n = Σ_(k=1and k=2p) ^n (((−1)^k )/k) +Σ_(k=1 and k=2p+1) ^n (((−1)^k )/k) =Σ_(p=1) ^([(n/2)]) (1/(2p)) − Σ_(p=0) ^([((n−1)/2)]) (1/(2p+1)) but Σ_(p=1) ^([(n/2)]) (1/(2p)) =(1/2) H_([(n/2)]) Σ_(p=0) ^([((n−1)/2)]) (1/(2p+1)) =1 +(1/3) + (1/5) +....+(1/(2[((n−1)/2)]+1)) =1+(1/2) +(1/3) +(1/4) +(1/5) +.....+(1/(2[((n−1)/2)])) +(1/(2[((n−1)/2)]+1)) −(1/2) −(1/4) −....−(1/(2[((n−1)/2)])) =H_(2[((n−1)/2)]+1) −(1/2) H_([((n−1)/2)]) ⇒ S_n = (1/2) H_([(n/2)]) −H_(2[((n−1)/2)]+1) +(1/2) H_([((n−1)/2)]) 2) we have S_(2n) =(1/2) H_n +(1/2) H_(n−1) −H_(2n−1) S_(2n) =(1/2){ ln(n) +γ +ln(n−1) +γ +o((1/n))} −ln(2n−1)−γ +o((1/n)) =(1/2)ln(n^2 −n)−ln(2n−1) +o((1/n)) =ln(((√(n^2 −n))/(2n−1))) +o((1/n)) ⇒lim_(n→+∞) S_(2n) = −ln(2) also S_(2n+1) =(1/2) H_n −H_(2n+1) +(1/2) H_n = H_n −H_(2n+1) = ln(n) +γ +o((1/n)) −ln(2n+1)−γ −o((1/n)) =ln((n/(2n+1))) +o((1/n)) ⇒lim_(n→+∞) S_(2n+1) = −ln(2) from that we can conclude that lim_(n→+∞) S_n =−ln(2).$
$S_{n} = \sum_{k = 1 a n d k = 2 p}^{n} \frac{{(- 1)}^{k}}{k} + \sum_{k = 1 a n d k = 2 p + 1}^{n} \frac{{(- 1)}^{k}}{k}$ $= \sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{2 p} - \sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{2 p + 1} b u t$ $\sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{2 p} = \frac{1}{2} H_{[\frac{n}{2}]}$ $\sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{2 p + 1} = 1 + \frac{1}{3} + \frac{1}{5} + . . . . + \frac{1}{2 [\frac{n - 1}{2}] + 1}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + . . . . . + \frac{1}{2 [\frac{n - 1}{2}]} + \frac{1}{2 [\frac{n - 1}{2}] + 1}$ $- \frac{1}{2} - \frac{1}{4} - . . . . - \frac{1}{2 [\frac{n - 1}{2}]}$ $= H_{2 [\frac{n - 1}{2}] + 1} - \frac{1}{2} H_{[\frac{n - 1}{2}]} \Rightarrow$ $S_{n} = \frac{1}{2} H_{[\frac{n}{2}]} - H_{2 [\frac{n - 1}{2}] + 1} + \frac{1}{2} H_{[\frac{n - 1}{2}]}$ $2) w e h a v e S_{2 n} = \frac{1}{2} H_{n} + \frac{1}{2} H_{n - 1} - H_{2 n - 1}$ $S_{2 n} = \frac{1}{2} {l n (n) + γ + l n (n - 1) + γ + o (\frac{1}{n})}$ $- l n (2 n - 1) - γ + o (\frac{1}{n})$ $= \frac{1}{2} l n (n^{2} - n) - l n (2 n - 1) + o (\frac{1}{n})$ $= l n (\frac{\sqrt{n^{2} - n}}{2 n - 1}) + o (\frac{1}{n}) \Rightarrow {l i m}_{n \to + \infty} S_{2 n} = - l n (2)$ $a l s o$ $S_{2 n + 1} = \frac{1}{2} H_{n} - H_{2 n + 1} + \frac{1}{2} H_{n}$ $= H_{n} - H_{2 n + 1} = l n (n) + γ + o (\frac{1}{n}) - l n (2 n + 1) - γ - o (\frac{1}{n})$ $= l n (\frac{n}{2 n + 1}) + o (\frac{1}{n}) \Rightarrow {l i m}_{n \to + \infty} S_{2 n + 1} = - l n (2)$ $f r o m t h a t w e c a n c o n c l u d e t h a t$ ${l i m}_{n \to + \infty} S_{n} = - l n (2) .$
Commented byabdo mathsup 649 cc last updated on 17/Jul/18

$3) w e h a v e S_{n}^{2} = {(\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k})}^{2}$ $= \sum_{k = 1}^{n} \frac{1}{k^{2}} + 2 \sum_{1 ⩽ i < j ⩽ n} \frac{{(- 1)}^{i} {(- 1)}^{j}}{i . j}$ $= \sum_{k = 1}^{n} \frac{1}{k^{2}} + 2 W_{n} \Rightarrow 2 W_{n} = S_{n}^{2} - \sum_{k = 1}^{n} \frac{1}{k^{2}}$ $\Rightarrow {l i m}_{n \to + \infty} W_{n} = \frac{1}{2} ({(- l n 2)}^{2} - \frac{π^{2}}{6})$ $= \frac{1}{2} ({(l n (2))}^{2} - \frac{π^{2}}{6}) .$
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