# Radius of Convergence of $\sum_{n = 0}^{\infty} \frac{(-1)^nn!x^n}{n^n}$

I'm taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn't provide explanations, and I'd really, really appreciate it if someone could explain why the answer is $e$.

which of the following is the radius of the convergence of the series $\large{\sum\limits_{n=0}^\infty\frac{(-1)^n n!x^n}{n^n}}$

$(A) \quad 0,\qquad (B) \quad\frac1e,\qquad (C)\quad1,\qquad (D) \quad e,\qquad (E) \infty$

Thank you all so much.

• Is this a "calculator problem" or is it in a non-calculator section? – Carser Apr 28 '16 at 18:15
• Calculator section - apologies for not mentioning that in my post. – Ozymandias Apr 28 '16 at 18:17

## 1 Answer

Denote $\sum a_n x^n$ the given series then

$$\left\vert\frac{a_{n+1}}{a_n}\right\vert=\left(1+\frac1n\right)^{-n}\xrightarrow{n\to\infty}\frac1e$$ so by the ratio test, the radius of convergence is $e$.

• I actually tried the ratio test, but the limit just gave me a radius of convergence of 1. It would be wonderful if you could show the steps you took to get there. – Ozymandias Apr 28 '16 at 18:18
• @Ozymandias You probably simplified the limit $\lim \left( \frac{n}{n+1} \right)^n$ to $1$ instead of $1/e$. It's a very common mistake to just assume that that ratio goes to one. – MCT Apr 28 '16 at 18:19
• But isn't the limit of (1+n)^n as n goes to infinity = e, not 1/e? – Ozymandias Apr 28 '16 at 18:37