Does $\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n$ converge? $$\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n$$

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*With the Root Test I get: $e\times \frac{1}{e}=1$, which doesn't determine whether it converges or not.


*With the Ratio Test I have to compute:
$$\lim_{n\to \infty}\left|\frac{\left(1+\frac{1}{n+1}\right)^{{(n+1)}^2}\left(\frac{1}{e}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n}\right|=\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{2n+1}\left(\frac{1}{e}\right)$$
\begin{align}
\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{2n+1}&=\exp\lim_{n\to \infty}\left(1+\frac{1}{n}-1\right)(2n+1)\\
&=\exp\lim_{n\to\infty}2+\frac{1}{n}\\
&=\exp2=e^2
\end{align}
$e^2\times \frac{1}{e}=e>1$, so it diverges
Is this correct? Im kinda new to this concept so I need a little help.
 A: You limit computations are not exactly correct even though they give the final good answer. However to exactly compute the limit:
\begin{align}
\lim_{n\to \infty} \left(1 + \frac1n\right)^{2n+1} &= \lim_{n\to \infty} \left(\left(1 + \frac1n\right)^n\right)^{2}\left(1+\frac1n\right) = e^2\times 1 = e^2
\end{align}
A: A necessary condition for convergence is that the $n^{th}$ term should go to zero. Let $\lim_{n \to \infty}{(1+1/n)}^{n^2}{(1/e)}^{n}=A.$ Then
$\implies \lim_{n \to \infty}{n^2}\log{(1+1/n)}+{n}\log{1/e}=\log A $
$\implies \lim_{n \to \infty}{n^2}{(1/n-1/{2n^2}+...)}-{n}=\log A  $
$\implies \log A=-1/2$
Contradiction, hence It is divergent. What you have done is also correct apart from what @youem mentioned.
A: You don't need either of these tests, because a necessary condition for convergence is that the general term tends to $0$.
Now, using Taylor-Young's expansion, we have
\begin{align}
\biggl(1+\frac1n\biggr)^{n^2}&=\mathrm e^{n^2\log\bigl(1+\frac1n\bigr)}=\mathrm e^{n^2\log\bigl(1+\frac1n\bigr)}\\
&=\mathrm e^{n^2\bigl(\frac1n-\frac1{2n^2}+o\bigl(\frac 1{n^2}\bigr)\bigr)}=\mathrm e^{n-\frac1{2}+o\bigl(1\bigr)}
\end{align}
so that $$\biggl(1+\frac1n\biggr)^{n^2}\frac1{\mathrm e^n}= \mathrm e^{-\frac1{2}+o\bigl(1\bigr)},$$
which tends to $\dfrac1{\sqrt{\mathrm e}}$.
A: As an alternative, using that $\log (1+x)\ge x-\frac {x^2}2$ we obtain
$$\left(1+\frac{1}{n}\right)^{n^2}=e^{n^2\log\left(1+\frac{1}{n}\right)}\ge e^{n-\frac12}=\frac{e^n}{\sqrt e}$$
and therefore
$$\left(1+\frac{1}{n}\right)^{n^2}\left(\frac{1}{e}\right)^n\ge \frac{e^n}{\sqrt e}\cdot \frac1{e^n}=\frac{1}{\sqrt e}$$
Refer also to the related

*

*Mean value theorem and inequality proving
