let-S-n-k-1-n-1-k-k-1-calculate-S-n-interms-of-H-n-2-find-lim-n-S-n-3-let-W-n-1-i-lt-j-n-1-i-j-i-j-prove-that-W-n-is-convergent-and-calculste-its-lim

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 by abdo mathsup 649 cc last updated on 17/Jul/18

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} + \dots . + \frac{1}{2 [\frac{n - 1}{2}] + 1}

= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots . . + \frac{1}{2 [\frac{n - 1}{2}]} + \frac{1}{2 [\frac{n - 1}{2}] + 1}

- \frac{1}{2} - \frac{1}{4} - \dots . - \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 by abdo 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}) .