let-u-n-n-n-1-1-n-1-find-the-nature-of-n-0-u-n-

Question Number 35057 by math khazana by abdo last updated on 14/May/18

l e t u_{n} = {(\frac{n}{n + 1})}^{\frac{1}{n}} - 1

f i n d t h e n a t u r e o f \sum_{n ⩾ 0} u_{n}

Commented by prof Abdo imad last updated on 16/May/18

w e h a v e u_{n} = {(1 - \frac{1}{n + 1})}^{n} - 1

= e^{n l n (1 - \frac{1}{n + 1})} - 1 b u t {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n}

\Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1} = - \sum_{n = 1}^{\infty} \frac{u^{n}}{n}

= - (u + \frac{u^{2}}{2} + \dots . .) \Rightarrow

l n (1 - \frac{1}{n + 1}) = - \frac{1}{n + 1} - \frac{1}{2 {(n + 1)}^{2}} + o (\frac{1}{{(n + 1)}^{3}}) \Rightarrow

n l n (1 - \frac{1}{n + 1}) = \frac{- n}{n + 1} - \frac{n}{{(n + 1)}^{2}} + o (\frac{1}{n^{2}}) \to - 1 (n \to + \infty)

\Rightarrow {l i m}_{n \to + \infty} u_{n} = \frac{1}{e} \neq 0 s o Σ u_{n} d i v e r g e s .