Does this serie $\sum\limits_{n=0}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}$ converge? 
Does this series $$\sum\limits_{n=0}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}$$ converge?.

Attempt : 
First I checked if $$u_n=\left(\dfrac{n}{n+1}\right)^{n^2}\underset{n\to +\infty}{\longrightarrow}0$$ because that is a necessary condition
$$u_n=\exp\bigg[n^2\bigg(\ln n-\ln(n+1)\bigg)\bigg]=\exp\bigg[-n^2\bigg(\ln (n+1)-\ln n\bigg)\bigg]$$
Equivalent of $\ln (n+1)-\ln n$ :
According to the mean value theorem, $\exists c \in ]n,n+1[$ such that
$\ln (n+1)-\ln n =f'(c)=\dfrac{1}{c}$ since $c\sim n$ so we have $\big(\ln(n+1)-\ln n\big) \sim \dfrac{1}{n} $ and $-n^2\bigg(\ln (n+1)-\ln n\bigg)\sim-n$
Thus :
$$\displaystyle \lim_{n\to +\infty} u_n =\lim_{n\to +\infty} e^{-n}=0$$
Now I need to find if this sequence is dominated by an another one, I already know it converges. But I am blocked...
 A: My own Idea Cauchy Rule
$$a_n=\left(\dfrac{n}{n+1}\right)^{n^2}$$
$$\sqrt[n]{a_n}=\left(1+\dfrac{1}{n}\right)^{-n} =e^{-\frac{\ln(1+\frac{1}{n})}{\frac{1}{n}}} \overset{h=\frac{1}{n}}{=}e^{-\frac{\ln(1+h)}{h}}\to e^{-1}<1$$
Hence your series converges.

OP attempt By equivalence 
  However, from OP Attempt, he should have shown instead that, 
  $$\lim_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}\Longleftrightarrow  a_n\sim e^{n} e^{-\frac{1}{2}}$$
  and therefore, 
  $$\sum\limits_{n=0}^{\infty}e^{-n}=\sum\limits_{n=0}^{\infty}(e^{-1})^{n}=\frac{1}{1-e^{-1}}$$
  this gives another prove of the convergence.
Proof of $\lim\limits_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}$ 

Setting $h=\frac{1}{n}$
$$a_n =\left(1+\dfrac{1}{n}\right)^{-n^2} =e^{-\frac{\ln(1+\frac{1}{n})}{\frac{1}{n^2}}}{=}e^{-\frac{\ln(1+h)}{h^2}}$$
Thus,
$$a_ne^{n} = e^{\frac{1}{h}-\frac{\ln(1+h)}{h^2}}=e^{-\frac{1}{2}+o(h)}$$
Since  $$\ln(1+h)= h -\frac{h^2}{2} +o(h^2) $$
Hence $$\lim_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}$$
A: Considering $$a_n=\left(\frac{n}{n+1}\right)^{n^2}\implies \log(a_n)=n^2\log\left(\frac{n}{n+1}\right)$$ $$\log\left(\frac{a_{n+1}}{a_n}\right)=(n+1)^2\log\left(\frac{n+1}{n+2}\right)-n^2\log\left(\frac{n}{n+1}\right)$$
$$\log\left(\frac{a_{n+1}}{a_n}\right)=(n+1)^2\log\left(1-\frac{1}{n+2}\right)-n^2\log\left(1-\frac{1}{n+1}\right)$$ COnsidering large $n$ and using Taylor series $$\log\left(\frac{a_{n+1}}{a_n}\right)=-1+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ and using $x=e^{\log(x)}$
$$\frac{a_{n+1}}{a_n}=\frac{1}{e}+\frac{1}{3 e n^2}+O\left(\frac{1}{n^3}\right)\sim \frac{1}{e}<1$$ then convergence
A: We know that 
$$
\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{-n} = \lim_{n\to \infty}\left( \frac{n}{n+1}\right)^{n} = \frac{1}{e}<\frac{1}{2}.
$$
Then, there is some $n_0$ such that $n\geq n_0$ implies that
$$
\left(\frac{n}{n+1} \right)^{n} < \frac{2}{3} \implies \left(\frac{n}{n+1} \right)^{n^2} < \left(\frac{2}{3}\right)^{n}.
$$
Therefore, your sum can be bounded by
$$
\sum_{n\geq 1}\left(\frac{n}{n+1} \right)^{n^2} < \sum_{n< n_0}\left(\frac{n}{n+1} \right)^{n^2} + \sum_{n\geq n_0}\left(\frac{2}{3}\right)^{n} < \infty.
$$
