The sum is
$ \sum _{n=1}^{\infty }2^n\left(\frac{n}{n+1}\right)^{n^2}\: $
I started by studying the asymptotic behavior
$ 2^n\left(\frac{n}{n+1}\right)^{n^2}\:=2^n\left[\left(1+\frac{1}{n}\right)^{n\:}\right]^{-n} $
So if the following sum is convergent
$ \sum _{n=1}^{\infty }{2^n}/e^{n} $
I suppose I should reason that the sum under question is also convergent (am I right about this?), I guess this is just for simplification.
Then I introduce $ \varepsilon >0 $, and as I see should prove that there exists such $ N<n $ that following thing holds for all $n$ and $p$
$ \left|{2^n}/{e^n}+{2^{n+1}}/{e^{n+1}}+...+{2^{n+p}}/{2^{n+p}}\:\right|<\varepsilon $
The least thing I can do from there on is this
$ {2^n}{e^{-n}}\left|1+{2}/{e}+...+{2^p}/{2^p}\:\right|<\varepsilon $
How do I prove such $N$ exists?