# Determine the radius of convergence of the power series $\sum \frac{x^n}{n^n}$

Determine the radius of convergence of the power series $\sum \frac{x^n}{n^n}$ and interval of convergence.

My idea:

Using ratio test take $a_n=\frac{x^n}{n^n}$

Then $a_{n+1}=\frac{x^{n+1}}{(n+1)^{n+1}}$

Now $\frac{a_{n+1}}{a_n}=\frac{x^{n+1}}{(n+1)^{n+1}}\times \frac{n^n}{x^n}$

How should I proceed next?

## 3 Answers

HINT:

$$\frac{a_{n+1}}{a_n}=\left(\frac{x^{n+1}}{(n+1)^{n+1}}\right) \left(\frac{n^n}{x^n}\right)=x\left(\frac{1}{(n+1)\underbrace{\color{blue}{\left(1+\frac1n\right)^n}}_{\to e}}\right)\to 0$$

$(|a_n|)^{1/n} = \frac{|x|}{n}$. $\text{lim sup}_{n \to \infty} (|a_n|)^{1/n} = 0$. Hence radius of convergence is $+\infty$. It can also be proved by comparing with $\sum \frac{x^n}{n!}$.

Let $m=2\lceil x\rceil$.

Then

$$\sum_{n=1}^\infty\frac{x^n}{n^n}=\sum_{n=1}^{m-1}\frac{x^n}{n^n}+\sum_{n=m}^\infty\frac{x^n}{n^n}<\sum_{n=1}^{m-1}\frac{x^n}{n^n}+\sum_{n=m}^\infty\frac1{2^n},$$ which converges.