TITLE: Prove divergence of $\sum_{n=1}^{\infty} \left( 1- \frac{\log n}n \right)^n$
QUESTION [5 upvotes]: I try to prove that this series diverges, the suggestion of the book is to show that $${\displaystyle \lim_{n \to \infty}n\left( 1- \dfrac{\log n}{n} \right)^{n}} =1$$
For this, my attempt was to try to see that
$$
{\displaystyle \lim_{n \to \infty} \log \left[ n\left( 1- \dfrac{\log n}{n} \right)^{n} \right] }=0 $$
using that $\log x \leq x-1$ for $x=1-\dfrac{\log n}{n}$, already has a inequalities
$$\log n + n \log \left( 1-\dfrac{\log n}{n} \right)  \leq  \log n +n\left[  \left( 1-\dfrac{\log n}{n} \right)  -1\right] =0$$
but i can not find a lower bound  $c_ {n}$ for $\log n + n \log \left( 1-\dfrac{\log n}{n} \right) $ such that $c_ {n}\rightarrow 0$

REPLY [2 votes]: $$ \left( 1- \dfrac{\log n}{n} \right)^{n}=e^{n\log\left( 1- \frac{\log n}{n}\right)}\sim_\infty e^{-\log n}=\frac{1}{n}$$
and the series $\sum \frac{1}{n}$ is divergent. Conclude.