cslculate-n-2-ln-1-1-n-n-

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

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)