find-n-2-1-1-n-2-

Question Number 33305 by abdo imad last updated on 14/Apr/18

f i n d \sum_{n = 2}^{\infty} (1 - \frac{1}{n^{2}})

Commented by prof Abdo imad last updated on 15/Apr/18

t h e Q i s f i n d \sum_{n = 2}^{\infty} l n (1 - \frac{1}{n^{2}})

Commented by abdo imad last updated on 19/Apr/18

l e t p u t S_{n} = \sum_{k = 2}^{n} l n (1 - \frac{1}{k^{2}})

w e h a v e S_{n} = l n (\prod_{k = 2}^{n} (1 - \frac{1}{k^{2}})) b u t w e h a v e

\prod_{k = 2}^{n} (1 - \frac{1}{k^{2}}) = \prod_{k = 2}^{n} \frac{k^{2} - 1}{k^{2}} = \prod_{k = 2}^{n} \frac{k - 1}{k} . \frac{k + 1}{k}

= (\frac{1}{2} \frac{2}{3} \frac{3}{4} \dots . \frac{n - 2}{n - 1} \frac{n - 1}{n}) (\frac{3}{2} \frac{4}{3} \frac{5}{4} \dots . \frac{n}{n - 1} \frac{n + 1}{n})

= \frac{1}{n} \frac{n + 1}{2} = \frac{n + 1}{2 n} \Rightarrow S_{n} = l n (\frac{n + 1}{2 n}) \Rightarrow {l i m}_{n \to + \infty} S_{n} = - l n (2)

s o \sum_{n = 2}^{\infty} l n (1 - \frac{1}{n^{2}}) = - l n (2) .