## Loser66 one year ago Find the radius of convergence of $$\sum_{n=1}^\infty \dfrac{n^{n+2}}{n!}z^n$$ I got e. Am I right?

1. SolomonZelman

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

2. misty1212

HI!!

3. misty1212

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

4. misty1212

ratio test works right?

5. SolomonZelman

$$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n+2}z^n}{n!} }$$ $$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n}n^2z^n}{n!} }$$ 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.

6. SolomonZelman

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

7. misty1212

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

8. misty1212

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

9. SolomonZelman

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

10. SolomonZelman

same thing:)

11. Loser66

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.

12. SolomonZelman

Anytime....

13. SolomonZelman

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

14. SolomonZelman

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