Find the interval of convergence of the following series $\sum_{n=1}^{\infty}\frac{n^n x^n}{n!}$

Question

Find the interval of convergence of the following series $$\sum_{n=1}^{\infty}\frac{n^n x^n}{n!}.$$

I've found the out that the series converges absolutely for $|x|<1/e$ but diverges for $|x|>1/e.$ So, when $x=1/e,$ we have

$$\sum_{n=1}^{\infty}\left(\frac{n}{e }\right)^n\frac{1}{ n!}.$$ Also, when $x=-1/e,$ we have

$$\sum_{n=1}^{\infty}(-1)^n \left(\frac{n}{e }\right)^n\frac{1}{ n!}.$$

My question is: how do I proceed from here?

Answer 1

Note that $n\mapsto a_{n}\equiv(n/e)^{n}/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $\lim_{n\rightarrow\infty}a_{n}=0$. Therefore, by the alternating series test, $\sum(-1)^{n}a_{n}$ converges.

Answer 2

HINT

By Stirling approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

we have that

$$\left(\frac{n}{e }\right)^n\frac{1}{ n!}\sim \frac1{\sqrt{2 \pi n}}$$

Answer 3

Stirling's approximation says that$${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}$$therefore the first sum diverges also the 2nd sum converges since $$\text{2nd sum}\sim-1+\sum_{n=1}^{\infty}\dfrac{(-1)^n}{\sqrt n}{\\=-1+\sum_{k=1}^{\infty}\dfrac{1}{\sqrt {2k}}-\dfrac{1}{\sqrt {2k+1}}\\=-1+\sum_{k=1}^{\infty}\dfrac{\sqrt {2k+1}-\sqrt {2k}}{\sqrt {2k}\sqrt {2k+1}}\\=-1+\sum_{k=1}^{\infty}\dfrac{1}{\sqrt {2k}\sqrt {2k+1}}\dfrac{1}{\sqrt {2k+1}+\sqrt {2k}}\\<-1+\sum_{k=1}^{\infty}\dfrac{1}{2k\sqrt{2k+1}}<\infty}$$which is obviously convergent

Answer 4

The radius of convergence of $\sum_{n=0}^\infty a_nx^n$ is given by $R = \lim_{n\to\infty} \left|\frac{a_{n}}{a_{n+1}}\right|$, if this limit exists.

We have

$$\left|\frac{\frac{n^n}{n!}}{\frac{(n+1)^{n+1}}{(n+1)!}}\right| = \frac{n^n}{(n+1)^n} = \frac1{\left(1+\frac1n\right)^n} \xrightarrow{n\to\infty} \frac1e$$

so $R = \frac1e$, as you predicted.