Convergence of $\sum_{n=1}^\infty \dfrac{n\log n}{e^n}?$ How to test the convergence of $\sum_{n=1}^\infty \dfrac{n\log n}{e^n}?$
 A: As usual, I love D'Alemberts ratio test:
$$a_n:=\frac{n\log n}{e^n}\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{(n+1)\log(n+1)}{e^{n+1}}\frac{e^n}{n\log n}=$$
$$=\frac{n+1}{n}\frac{\log(n+1)}{\log n}\frac{1}{e}\xrightarrow[n\to\infty]{}\frac{1}{e}<1$$
and thus the series converges.
A: You can easily show that
$$
\frac{n\log n}{e^n}=O\left(\frac{1}{n^2}\right).
$$
All you have to do is check that
$$
\lim_{n\rightarrow +\infty}n^2 \cdot\frac{n\log n}{e^n}=0.
$$
Then conclude the series converges (absolutely, of course) by comparison with the Riemann series $\sum_{n\geq 1}  \frac{1}{n^2}$.
A: Another Hint:
$$\lim_{n\to\infty}n^{2}\times\frac{n\log(n)}{\exp(n)}=0<\infty$$ so it converges.
A: First note that $e^n > n^4$ for $n$ large enough. That is, first prove that there is a $K$ such that $e^n > n^4$ for all $n\geq K$.
Then you have for $n\geq K$ also that $\log(n) < n$ and so
$$
\frac{n\log(n)}{e^n} < \frac{n\log(n)}{n^4} < \frac{n^2}{n^4} = \frac{1}{n^2}.
$$
Hence you can compare the sequence to $\sum\frac{1}{n^2}$.
A: The ratio test is probably the best tool to use in this case (and in many other examples), but you may also use the integral test.
Let $f(x)=x\log x/e^x$. Clearly $f(x)$ is positive for all $x > 1$. Moreover $f(x)$ is decreasing at least on $[e,\infty)$. Indeed,
$$f'(x) = \frac{1+(1-x)\log(x)}{e^x}$$
so if $x\geq e$, then the numerator of $f'(x)$ is less than $1+(1-e)<0$. Hence $f'(x)<0$ for $x\geq e$, and $f$ is decreasing. Hence, we may apply the integral test on the series $\sum_{n=1}^\infty f(n)$.
The series $\sum_{n=1}^\infty f(n)$ converges if and only if $\sum_{n=3}^\infty f(n)$ converges and this series, by the integral test, converges if and only if $\int_3^\infty f(x)dx$ converges. Now:
$$\int_3^\infty f(x) dx = \lim_{N\to \infty} \int_3^N \frac{x\log x}{e^x} dx \leq \lim_{N\to \infty} \int_3^N \frac{x^2}{e^x} dx = \lim_{N\to\infty} [-e^{-x}(x^2+2x+2)]_3^N$$
$$ = \lim_{N\to\infty} -\frac{N^2+2N+2}{e^N}+\frac{17}{e^3}=\frac{17}{e^3}.$$
Since the improper integral converges, we conclude that the series $\sum_{n=1}^\infty f(n)$ converges as well.
