Does this power series $\sum_{n=0}^{\infty} \frac{n^n}{(n!)^2}x^n$ converge for all $x$? Does this power series $\sum_{n=0}^{\infty} \frac{n^n}{(n!)^2}x^n$ converge for all $x$? It was told to me that the series does converge for all $x$, however I have investigated with a computer behavior of $\frac{n^n}{(n!)^2}100^n$ and it looks like that the sequence diverges to $\infty$ (thus the series can not converge at least for $x=100$). 
 A: Denote $\sum a_n x^n$ this series. We have
$$\frac{a_{n+1}}{a_n}=\left(1+\frac1n\right)^n\frac1{n+1}\xrightarrow{n\to\infty} e\times0=0$$
so by the ratio test the radius of convergence is $R=\frac10=+\infty$.
A: If we try the ratio test, 
$$
\frac{\frac{(n+1)^{n+1}x^{n+1}}{((n+1)!)^2}}{\frac{n^nx^n}{(n!)^2}}
=\frac{(n+1)^{n+1}x}{n^n(n+1)^2}=\left(1+\frac1n\right)^n\frac{x}{n+1}\to0.
$$
As $n\to\infty$. So the series converges for all $x$. 
A: You're looking at
$$
\sum_{n=0}^{\infty}\left(\frac{n x}{(n!)^{2/n}}\right)^n.
$$
If the expression in parentheses ever drops below $1$ and stays there, then the series converges geometrically.  If you know that $(n!)^{1/n}\sim n/e$, then you can see that this will eventually happen for any $x$, since the expression in parentheses is asymptotic to $xe^2/n$... but until that happens, the series will appear to be diverging.  Numerically, for $x=100$ the terms are growing until $n\approx 270$, by which point they are very large ($\sim 7 \times 10^{114}$).  The terms then shrink, but slowly, because the expression in parentheses is shrinking but not yet less than $1$.  This doesn't happen until $n\approx 730$, after which the remainder series converges rapidly.
A: You will get a few answers that use the Sterling formula for the asymptotic behavior of $n!$. You don't need that though. Observe that 2/3 of the factors in $n!$ are greater than $n/3$, and therefore $n! > (n/3) ^ {(2n/3)}$. Use this in your computation of the radius of convergence for your series.
A: Yet another approach, using only elementary facts: (written here for the sake of getting an alternative proof)
You can write $\lvert x\rvert n \leq x^2 \mathbb{1}_{\{\lvert x\rvert > n\}} + n^2 \mathbb{1}_{\{\lvert x\rvert \leq n\}}$ so that 
$$
0 \leq \sum_{n=0}^\infty \frac{(\lvert x\rvert n)^n}{n!^2}  \leq 
\sum_{n=0}^\infty \left(\frac{x^2}{n!}\right)^n \mathbb{1}_{\{\lvert x\rvert > n\}} + \sum_{n=0}^\infty \left(\frac{n^2}{n!}\right)^n \mathbb{1}_{\{\lvert x\rvert \leq n\}}
\leq 
\sum_{n=0}^\infty \left(\frac{x^2}{n!}\right)^n + \sum_{n=0}^\infty \left(\frac{n^2}{n!}\right)^n
$$
Now, the RHS:


*

*The first term converges for all $x\in\mathbb{R}$ (compare it to the series defining the exponential, for instance: it is the square of its general term; or use the ratio test); 

*the second does not depend on $x$, and is easily seen to converge (for instance, $\frac{n^2}{n!} < \frac{2}{3}$ for $n \geq 4$).

