Does the following series $\sum_{n=1}^{\infty} \frac{n}{e^{n+1}}$ converge or diverge? $\sum_{n=1}^{\infty} \frac{n}{e^{n+1}}$
I am trying to evaluate the series via the integral comparison test and I feel like I was wondering if my work is correct:
$$\begin{align*}
\int_{1}^{\infty} x(e^{-x+1})dx = \lim_{T\to\infty} \int_{1}^{T}x(e^{-x+1})dx = e^{-T+1} - e^{-1}(2) = \frac{2}{e}
\end{align*}$$
 so I think the series is convergent.
I applied integration by parts on the indefinite integral with $u = x, du = 1, v = -e^{-x+1}, dv = e^{-x+1}$
$$\begin{align*}
\int x(e^{-x+1})dx &= -xe^{-x+1} - \int e^{-x+1}dx \\
&= -xe^{-x+1} + e^{-x+1} \\
&= e^{-x+1}(x+1)
\end{align*}$$
Did I apply do the integration correctly?
 A: If $|r|<1$, we have that $\sum_{n=1}^\infty r^n=\frac{1}{1-r}$ (usual geometric series). Differentiating both sides with respect to $r$, and subsequently multiplying by $r^2$,  we get 
$$\sum_{n=1}^\infty nr^{n+1}=\frac{r^2}{(1-r)^2}.$$
Put $r=\frac1e$ and you get that it converges to $\boxed{\tfrac{1}{(e-1)^2}}$.
A: Use the root test. Note that
$$
\left(\frac{n}{e^{n+1}}\right)^{1/n}=\frac{n^{1/n}}{e^{1+n^{-1}}}\to\frac{1}{e}<1
$$
and hence the series converges. We have used the well known limit $n^{1/n}\to 1$ as $n\to \infty$.
A: I’m not sure with your calculation for the integral, as an alternative we can simply use limit comparison test, that is
$$\frac{\frac{n}{e^{n+1}}}{\frac1{n^2}}\to 0$$
A: The ratio test seems to be easily applicable to this series:
$$
\begin{align}
\lim_{n\to\infty}\frac{a_{n+1}}{a_n}
&=\lim_{n\to\infty}\frac{\frac{n+1}{e^{n+2}}}{\frac{n}{e^{n+1}}}\\
&=\frac1e\lim_{n\to\infty}\frac{n+1}n\\
&=\frac1e\\[6pt]
&\lt1
\end{align}
$$
so the series converges.
A: $$\sum_{n\geq 0}z^n = \frac{1}{1-z} $$
defines a holomorphic function in the region $|z|<1$. The application of the operator $\frac{d}{dz}$ or $z\cdot\frac{d}{dz}$ does not change the radius of convergence, hence for any $k\in\mathbb{N}$ and any $z$ such that $|z|<1$ we have that
$$ \sum_{n\geq 0} n^k z^n $$
is convergent. In your case $k=1$ and $z=\frac{1}{e}$.
