Find the radius of convergence of
$\sum_{n=1}^\infty \dfrac{n^{n+2}}{n!}z^n$
I got e. Am I right?

Question

Find the radius of convergence of 
$\sum_{n=1}^\infty \dfrac{n^{n+2}}{n!}z^n$ 
I got e. Am I right?

solomonzelman · Answer

it seems as though it is |z|<1/e if you think of stirling's approximation
z! = z^z • √(2πz) ÷ e^z

misty1212 · Answer

HI!!

misty1212 · Answer

you got it upside down i think 
oh, what @SolomonZelman  said

misty1212 · Answer

ratio test works right?

solomonzelman · Answer

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n+2}z^n}{n!} }$

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n}n^2z^n}{n!} }$

by Stirling's approximation,

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n}n^2e^nz^n}{n^n\sqrt{2\pi n}} }$

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^2e^nz^n}{\sqrt{2\pi n}} }$

$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^2e^n}{\sqrt{2\pi n}\left(\dfrac{1}{z}\right)^n} }$

So if:
|1/z|>e    --->   z<1/e

then the series will converge.

solomonzelman · Answer

do the ratio test tho, that might be off just a bit.

misty1212 · Answer

$$\frac{(n+1)^{n+3}\times n!}{(n+1)!n^{n+2}}=\frac{(n+1)^{n+2}}{n^{n+2}}$$

misty1212 · Answer

you do get $e$ as the limit, but that means $$|z|<\frac{1}{e}$$

solomonzelman · Answer

$\large\color{black}{ \displaystyle \lim_{n \to \infty} \frac{n!(n+1)^{n+3}z^{n+1}}{(n+1)!n^{n+2}z^n} }$

$\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{n!(n+1)^{n+3}}{(n+1)!n^{n+2}} }$

$\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{n!(n+1)^{n+2}}{n!\cdot n^{n+2}} }$

$\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{(n+1)^{n+2}}{n^{n+2}} }$

$\large\color{black}{ \displaystyle \lim_{n \to \infty} z\left(1+\frac{1}{n}\right)^{n+2}=ez}$

|ez|<1
|z|<1/e

solomonzelman · Answer

same thing:)

loser66 · Answer

I use Hamadard test, which is 
$a_n = \dfrac{n^{n+2}}{n!}$ 

then $\sqrt[n]|a_n| = \dfrac{\sqrt[n]{n^2n^n}}{\sqrt[n]n!}$
 
and we know that the denominator $\approx n/e $. Hence, it is

$\dfrac{n\sqrt[n] n^2}{(n/e)}= e\sqrt[n]n^2$ 
oh, yeah, I forgot to take $R = 1/ limsup \sqrt[n]|a_n|$ that is why I got e, but 1/e 
Thank you all.

solomonzelman · Answer

Anytime....

solomonzelman · Answer

Yeah, just a note that this denominator of ≈n/e is also a consequence of Stirling's approximation.

solomonzelman · Answer

in any case, it is *|z|<1/e* .....