How may I prove that the following summation converges or diverges?
I tried to prove that $a(n)$ doesn't go to zero (which proves that the summation diverges) but finding the limit was too hard.
$$\sum_{n=0}^{\infty} a(n) =$$ $$\sum_{n=0}^{\infty} \frac{(1+\frac{(-1)^{n}}{n})^{n^{2}}}{n \cdot e^n}$$
Edit I proved that it's smaller than: $$\sum_{n=0}^{\infty} \frac{2^{n^2}}{(n \cdot e^n)}$$ if I could prove that the latter converges then I solved it
When $n$ is even we have $$\ln a(n)=n^2\ln(1+1/n)-\ln n-n=$$ $$=n^2(1/n-1/2n^2+1/3n^3-...)-\ln n-n>$$ $$>n^2(1/n-1/2n^2)-\ln n-n=-1/2-\ln n.$$ So $a(n)>1/n\sqrt e$ when $n$ is even.
Every $a(n)$ is positive so $$\sum_na(n)\ge \sum_{n \;\text {even}}a(n)\ge$$ $$\ge\sum_{n \;\text {even}}1/n\sqrt e=\infty.$$