Series $\sum_{n=1}^{\infty}(1-\frac2n)^{-n^2}\frac1{e^{2n}}$

Question

How can I determine the convergence or divergence of this series ? The root test doesn't give information here. $$\sum_{n=1}^{\infty}\left(1-\frac2n\right)^{-n^2}\frac1{e^{2n}}$$

Answer 1

Hint. Consider the limit of the $n$-th term $$\lim_{n\to +\infty}a_n=\lim_{n\to +\infty}(1-2/n)^{-n^2}e^{-2n}= \lim_{n\to +\infty}\exp(-n^2\ln(1-2/n)-2n)$$ and show that it is not zero (but different from $1$). What may we conclude?

P.S. The index $n$ should be greater than $2$ otherwise $a_2$ is not defined.

Answer 2

Notice $e^x \ge 1 + x > 0$ for all $x > -1$. For $n \ge 2$, this leads to $$e^{-\frac2n} \ge 1 - \frac2n > 0 \quad\implies\quad e^{-2n} \ge \left(1-\frac2n\right)^{n^2} \quad\implies\quad a_n = \left(1-\frac2n\right)^{-n^2} e^{-2n} \ge 1 $$ Since $a_n$ doesn't converge to $0$, the series $\sum\limits_{n=3}^\infty a_n$ diverges.