find-n-1-ln-1-1-n-

Question Number 33304 by abdo imad last updated on 14/Apr/18

f i n d \sum_{n = 1}^{\infty} l n (1 + \frac{1}{n})

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

l e t p u t S_{n} = \sum_{k = 1}^{n} l n (1 + \frac{1}{k}) w e h a v e \sum_{n = 1}^{\infty} l n (1 + \frac{1}{n}) =_{n \to + \infty} l i m S_{n}

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

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

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