calculate-k-2-ln-1-1-k-2-

Question Number 28619 by abdo imad last updated on 27/Jan/18

c a l c u l a t e \sum_{k = 2}^{+ \infty} l n (1 - \frac{1}{k^{2}}) .

Commented by abdo imad last updated on 31/Jan/18

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

w e h a v e S = {l i m}_{n \to + \infty} S_{n} b u t

S_{n} = l n (\prod_{k = 2}^{n} (1 - \frac{1}{k^{2}})) b u t

\prod_{k = 2}^{n} (1 - \frac{1}{k^{2}}) = \prod_{k = 2}^{n} (\frac{k^{2} - 1}{k^{2}}) \Rightarrow l n (\prod_{k = 2}^{n} (1 - \frac{1}{k^{2}}))

= \sum_{k = 2}^{n} l n (k - 1) + \sum_{k = 2}^{n} l n (k + 1) - 2 \sum_{2}^{n} l n (k)

= \sum_{k = 1}^{n - 1} l n (k) + \sum_{k = 3}^{n + 1} l n (k) - 2 \sum_{k = 2}^{n} l n (k)

= \sum_{k = 2}^{n} l n k - l n (n) + \sum_{k = 2}^{n} l n (k) - l n 2 + l n (n + 1) - 2 \sum_{k = 2}^{n} l n (k)

= - l n (n) + l n (n + 1) - l n 2 = l n (\frac{n + 1}{n}) - l n (2)

s o {l i m}_{n \to + \infty} S_{n} = - l n (2) = S .