Question

Find the radius of convergence of the series:

Answer 1

\[\large \sum_{n=1}^{\infty} \frac{n!x^n}{n^n}\]

What I did was apply the root test, so:
\[\huge \lim_{n \rightarrow \infty}\sqrt[n]{\frac{n!x^n}{n^n}}= x \lim_{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n}\]

I applied L'Hospital's rule and determined that the limit is 0, which means the radius should be \(\infty\), but that's not the right answer =/

Answer 2

Iʻd ask @Hero

Answer 3

Oh wait, that just means that it's convergent, right? Since L<0. It's been a while since I've taken Calculus 2 >_< Taking differential equations right now..

Answer 4

L<1*

Answer 5

http://www.wolframalpha.com/input/?i=Lim [%28%282+Pi+n%29^%281%2F%282n%29%29%29%2Fe%2Cn-%3EInfinity]

\[
\lim_{x\to\infty}\frac{2\pi n^{\frac{1}{2n}}}{e}=\frac{1}{e}
\]
Don't ask me why.

Answer 6

Put this into WolframAlpha.

Lim[((2 Pi n)^(1/(2n)))/e,n->Infinity]

Answer 7

That doesn't make sense to me though. I don't necessarily want the answer -- I want to know how to solve it.

Answer 8

use ratio test

Answer 9

@Zarkon Shouldn't they give me the same answer?

Answer 10

yes...ratio test is the method to use for this problem though

Answer 11

otherwise you need to use Stirling's approximation

Answer 12

\[
\begin{align*}
&\phantom{{}={}}\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}\\
&=\lim_{n\to\infty}\sqrt[n]{\frac{\sqrt{2\pi n}(\frac{n}{e})^n}{n^n}}\\
&=\lim_{n\to\infty}\frac{(2\pi n)^\frac{1}{2n}(\frac{n}{e})}{n}\\
&=\lim_{n\to\infty}\frac{(2\pi n)^\frac{1}{2n}}{e}\\
&=\frac{1}{e}\lim_{n\to\infty}n^\frac{1}{2n}
\end{align*}
\]
Take logarithm of that limit than solve it.

Answer 13

I still got 0 =/

`Ratio Test`:
\[\large \lim_{n \rightarrow \infty}\frac{ (n+1)!x^{n+1} }{ (n+1)^{n+1} }* \frac{ n^n }{ n!x^n }=\lim_{n \rightarrow \infty}\frac{ n }{ (n+1)^n }\]\[\large L'Hospital's \rightarrow \lim_{n \rightarrow \infty}\frac{ 1 }{ n(n+1)^{n-1} }=0\]

Answer 14

that is because you did it wrong

Answer 15

Oh, where did I mess up?

Answer 16

\[\frac{ (n+1)!x^{n+1} }{ (n+1)^{n+1} }\frac{ n^n }{ n!x^n }=\frac{(n+1)xn^n}{(n+1)^{n+1}}=\frac{xn^n}{(n+1)^{n}}\]

Answer 17

Ah yeah I left out the exponent of n on the top..
But how would I take the limit of that then?

Answer 18

\[\frac{xn^n}{(n+1)^{n}}=\frac{x}{\frac{(n+1)^n}{n^n}}=\frac{x}{\left(\frac{n+1}{n}\right)^n}=\frac{x}{\left(1+\frac{1}{n}\right)^n}\]

Answer 19

\[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e\]

Answer 20

Why does that equal e? Wouldn't it be 1 then? 1/infinity (approaching it) would be 0. so (1+0)^infinity would be 1?

Answer 21

http://mathworld.wolfram.com/e.html

see equation (3). it is a definition of e

Answer 22

Ah okay, I was never taught that. Definitely good to know now though. Thanks!!