Convergence test for: $\sum\limits_{n = 1}^{\infty} \frac{n}{e^n}$ I know, for example, that the series:
$$\sum\limits_{n= 1}^{\infty} \frac{1}{e^n}$$
is a geometric series of the form $\sum\limits_{n = 1}^{\infty} k x^n$, where $k = 1$ and $x = \frac{1}{e}$ and it is convergent1. But when I have, as in my case:
$$\sum\limits_{n = 1}^{\infty} \frac{n}{e^n}$$
a ratio of functions I'm struggling to find a method to find whether the series is convergent/ divergent and to find a value in the former case. Thus the following question arises:
What series convergence method to use in case of ratio of functions?

1. And its value is: $\frac{k}{1 - x} - 0^{th}term$
 A: Let $a_n = \frac{n}{e^n}$. Then by ratio test you get that:
$$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right| = \frac{1}{e} < 1.$$
So you can conclude that the series is convergent. Hope that it helps you :)
A: Hint:
$$\frac{x}{(1-x)^2}=\sum_{n=0}^{\infty }nx^n$$ for $|x|<1$
A: You can also use the root test which is stronger than the ratio test. Using the well-known limit  $$\lim_{n\to \infty} n^{1/n}=1$$ we have that $$\left(\frac{n}{e^n}\right)^{1/n}=\frac{1}{e}n^{1/n} \longrightarrow \frac{1}{e}<1$$ so that the series converges.
A: $\sum \frac{n}{e^n} \leq \sum \frac{n}{n^3} = \sum \frac{1}{n^2}<\infty.$
A: If $\{a_n\}$ is a nonincreasing sequence of nonnegative numbers, then $\sum_n a_n$ converges iff $\sum_n 2^n a_{2^n} $ converges. (This is known as the Cauchy condensation test, and can be used to show that $\sum_n \frac1{n^p}$ converges iff $p>1$). Applying this test, we have
$$2^n\left(\frac{2^n}{e^{2^n}}\right) = \frac{4^n}{(e^2)^n} = \left(\frac 4{e^2}\right)^n. $$
So the series converges (as $4/e^2<1$).
A: By the integral test,
$$\int_0^\infty\frac x{e^x}dx=-\left.\frac x{e^x}\right|_0^\infty+\int_0^\infty\frac{dx}{e^x}=-\left.\frac 1{e^x}\right|_0^\infty=1.$$
A: Ratio test proves it converges. But comparison test is also a powerful tool.
In this example, 
e^x >x^3   For x>5
So using that fact and p test, it Can be shown that the series converges
