Convergence test for: $\sum\limits_{n = 1}^{\infty} \frac{n}{e^n}$ [duplicate]

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: $\frac{k}{1 - x} - 0^{th}term$

marked as duplicate by Alex M., Henrik, Lee Mosher, Shailesh, Parcly TaxelSep 10 '16 at 3:05

• The ratio test is convenient when it works. The root test works more often, but is sometimes harder to apply. In this example, both work and are easy. – Daniel Fischer Sep 9 '16 at 15:03
• The ratio or root test will demonstrate that this series converges absolutely – Joel Sep 9 '16 at 15:04
• Do you know what $\sum_{n = 1}^\infty n x^n$ is? Substitute $x = \frac{1}{\mathrm{e}}$ afterwards. – Ritz Sep 9 '16 at 15:06
• You need the denominator to grow sufficiently much faster than the numerator to have convergence. But in general, if you have somewhat complicated (positive) functions $f$ and $g$, looking at the asymptotic behaviour helps to determine whether $\sum \frac{f(n)}{g(n)}$ converges. Replacing $f$ and $g$ by asymptotically equivalent functions preserves convergence/divergence [that is the limit comparison test]. Note however that this doesn't (generally) hold if $f$ or $g$ change sign. – Daniel Fischer Sep 9 '16 at 15:20
• @Ziezi: theintegral and ratio tests will work. Try them. – Yves Daoust Sep 9 '16 at 15:21

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 :)

Hint:

$$\frac{x}{(1-x)^2}=\sum_{n=0}^{\infty }nx^n$$ for $|x|<1$

• Most probably power series are still pretty far away from the OP's assumed and guessed level, as this is a rather elementary question in number series. – DonAntonio Sep 9 '16 at 15:14

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.

$\sum \frac{n}{e^n} \leq \sum \frac{n}{n^3} = \sum \frac{1}{n^2}<\infty.$

• So, the last is a p-series with $p = 2 > 1$ - so it converges? – Ziezi Sep 9 '16 at 15:13
• @Ziezi Yes, yet the inequality used, $\;e^n\ge n^3\;$ may not be as immediate to catch...but if you did then you're done. – DonAntonio Sep 9 '16 at 15:16

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 $4/e^2<1$).

By the integral test,

$$\int_0^\infty\frac x{e^x}dx=-\left.\frac x{e^x}\right|_0^\infty+\int_0^\infty\frac{dx}{e^x}=-\left.\frac 1{e^x}\right|_0^\infty=1.$$

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