# Interval of convergence

1. Dec 17, 2017

### Silviu

1. The problem statement, all variables and given/known data
Find the interval of convergence of: $\sum\frac{n^n}{n!}z^n$

2. Relevant equations

3. The attempt at a solution
I obtained that the radius of convergence is $1/e$ but I am not sure what to do at the end points. For $z=1/e$ I would have $\sum{n^n}{n!e^n}$.
Mod edit: I think you mean $\sum \frac{n^n}{n!e^n}$.
Using Stirling formula I would obtain an approximation of the form $\sum \frac{1}{\sqrt{2 \pi n}}$, which would go to infinity. However I am not sure how to make it formal, as this approximation is not a lower, but an upper bound for n!. And how should I proceed for -1/e? Thank you!

Last edited by a moderator: Dec 17, 2017
2. Dec 17, 2017

### Ray Vickson

As shown in Volume I of Fuller's probability book, we have rigorous lower and upper bounds:
$$\sqrt{2 \pi n} \, n^n e^{-n} < n! < \sqrt{2 \pi n} \, n^n e^{-n + 1/(12 n)}$$
for all integers $n \geq 1$. For example, for $n = 1$ and $n = 2$ these give $0.922137 < 1! < 1.00227$ and $1.919004 < 2! < 2.00065$. Note that the upper bound is a very much better approximation to $n!$ than the simple Stirling formula, and is quite accurate even for $n$ as small as $1$ or $2$ (getting ever more accurate with larger $n$).

Feller's book may be found in
https://www.amazon.ca/Introduction-Probability-Theory-Applications-Vol/dp/0471257087

A free, on-line version is at
https://eclass.uop.gr/modules/docum...μπληρωματικά έντυπα/FellerProbabilityVol1.pdf

Look, in particular, in Chapter II, pp. 52--54.

3. Dec 18, 2017

### Silviu

Thank you, but this still doesn't answer my question...

4. Dec 18, 2017

### Ray Vickson

Yes, it does. You needed an upper bound (on $n!$) and you got one. What you do with it is up to you. It is against PF rules for a helper to solve a problem, except, possibly, to show an easier or shorter way to deal with a correct solution that has already been posted.

Last edited: Dec 18, 2017