<s>[INST] Proving convergence of $ \sum_{n=1}^{\infty} \left({n \over n+1}\right)^{n^2} $ $$ \sum_{n=1}^{\infty} \left({n \over n+1}\right)^{n^2} $$
We have :
$$ \left({n \over n+1}\right)^n = \left({1 \over 1+{1\over n}}\right)^n = 
{1 \over  \left( 1+ {1 \over n}\right)^n}$$ 
So,
$$ \lim_{n \rightarrow \infty} \left({n \over n+1}\right)^n = {1 \over e} \lt 1 $$
And the series converges by Cauchy criterion.
Is my proof correct?
 [/INST] By comparing this series with respect to this geometric series we found that, your series converge to the a number less than $1$:
$$\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}=\frac{1}{2}+(\frac{2}{3})^4+(\frac{3}{4})^9+(\frac{4}{5})^{16}...< \frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+(\frac{1}{2})^4+...=1$$
If we could a more restricted upper bound to series, we reach to more accurate number. 
To reach a lower bound I use similar way:
$$\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}=\frac{1}{2}+(\frac{2}{3})^4+(\frac{3}{4})^9+(\frac{4}{5})^{16}+...> \frac{1}{2}+(\frac{1}{2})^3+(\frac{1}{2})^5++(\frac{1}{2})^7...=\frac{2}{3}$$
So:
$$\frac{2}{3}<\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}<1$$
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