TITLE: Does the series $\sum\limits_{n=1}^{\infty }\left ( 1-\frac{\ln(n)}{n} \right )^{2n}$ diverge?
QUESTION [9 upvotes]: could anyone help me figure out whether this infinite series
$$\sum_{n=1}^{\infty }\left ( 1-\frac{\ln(n)}{n} \right )^{2n}$$
diverges?
I've tried using Cauchy's and d'Alembert's limit tests but both gave the result 1. I've also tried the necessary condition for convergence, but 
$$\lim_{n\rightarrow \infty }\left ( 1-\frac{\ln(n)}{n} \right )^{2n}=0$$

REPLY [3 votes]: Using Cauchy Condensation, we get the series. 
$$\sum_{n=1}^\infty 2^nf(2^n) = \sum_{n=1}^{\infty}\left( 1 - \frac{n \log 2}{2^n}\right)^{2^{n+1}} 2^n \tag{1}$$
Here we can use root test, 
$$\lim_{n\to\infty} \sqrt[n]{\left( 1 - \frac{n \log 2}{2^n}\right)^{2^{n+1}} 2^n} = \frac 1 2 \tag{2}$$
to argue that $(1)$ converges which implies the convergence of out original series.
To evaluate the limit $(2)$, 
$$2 \lim_{n\to\infty} \left( 1 - \frac{n \log 2}{2^n}\right)^{\frac{2^n}{n \log 2}2 \log 2} = 2 e^{- 2\log 2} = \frac 1 2$$