# How to evaluate $\lim\limits_{n \to\infty} \left|\frac{(n+1)^n x^{(n+1)}}{n^n}\right|$?

I'm trying to find the radius of convergence for the series $\sum_{n=0}^{\infty} \frac{n^n}{n!}x^n$ and have used Wolfram Alpha to find that it is $|x| < \frac{1}{e}$ and am trying to show that myself.

I'm trying to show it using the limit. So I've simplified the limit to $\lim\limits_{n \to\infty} \left|\frac{(n+1)^n x^{(n+1)}}{n^n}\right|$ and I'm not sure where to go from there.

Any help or hints would be appreciated.

• You rather obtain $\lim_{n \to\infty} \left|\frac{(n+1)^n}{n^n} x \right|$, and see my answer below. Thanks. – Olivier Oloa Aug 1 '15 at 22:51
• Note that for a series $\sum a_n x^n$ the radius of convergence is given by $\frac{1}{R} = \lim\limits_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim\limits_{n\to\infty}\left|\frac{(n+1)^{n}}{n^n}\right|$. Alternatively, we have that a series $\sum b_n$ converges for $\lim\limits_{n\to\infty}\left|\frac{b_{n+1}}{b_n}\right| < 1$, which in this case gives $\lim\limits_{n\to\infty}\left|\frac{(n+1)^{n} x}{n^n}\right| < 1$ . – Winther Aug 1 '15 at 22:56

By ratio test $$lim_{n \to \infty }|\frac{a_{n+1}}{a_n}| =\\lim_{n \to \infty }|\frac{\frac{x^{n+1}(n+1)^{n+1}}{(n+1)!}}{\frac{x^{n}(n)^{n}}{(n)!}}| =\\ lim_{n \to \infty } |x\frac{(n+1)^n}{n^n}|=\\lim_{n \to \infty } |xe|=l \left\{\begin{matrix} {\color{Red} {l <1}} & convergence\\ l=1 & another \space test\\ l>1 & divergence \end{matrix}\right. \\ {\color{Red}{|ex|<1 \\|x|<\frac{1}{e}} }$$

Hint. One may recall that $$\lim_{n \to \infty}\left(1+\frac1n\right)^n=e.$$

• $$lim_{n \to\infty} |\frac{(n+1)^n x^{(n+1)}}{n^n}|=\\lim_{n \to\infty} |\frac{(\frac{n+1}{n})^n x^{(n+1)}}{1}|\\ lim_{n \to\infty} |{(1+\frac{1}{n})^n x^{(n+1)}}|=\\e *lim_{n \to\infty} | x^{(n+1)}|$$ – Khosrotash Aug 1 '15 at 22:51
• @Khosrotash Please, see my comment above. Thanks. – Olivier Oloa Aug 1 '15 at 22:52
• $$\frac{1}{R}=lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|$$ – Khosrotash Aug 1 '15 at 22:55
• Thanks guys! I ended up with what @Khosrotash got in his first comment, using the hint that Olivier Oloa had... But I don't think I can just say that the $lim_{n \to \infty} |x^{(n+1)}|$ will give me x...which I think is what I would need in order to continue. That way I would have that the limit = $|e*x|$ This way I can rearrange the inequality of the limit being less than 1. Is my line of thinking correct? – Rebecca Aug 1 '15 at 23:00
• @Rebecca Please have a look to winther comment. Thanks. – Olivier Oloa Aug 1 '15 at 23:02

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

to see that: $$\frac{n^n}{n!} \sim \frac{e^n n^n}{\sqrt{2 \pi n} n^n} = \frac{e^n}{\sqrt{2 \pi n}}$$

You need $r$ such that:

$$\lim_{n \to \infty} \frac{n^n r^n}{n!} = 1$$

and that is seen to be $1/e$.