# For what values of x $\sum_{n=1}^{\infty} \frac{(x-4)^{n^{2}}}{n!}$ converges?

$$\sum_{n=1}^{\infty} \frac{(x-4)^{n^{2}}}{n!}$$ I've tried root test but I couldnt solve

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Could you show us where you got stuck using the root test? – Shaun Ault Feb 20 '13 at 14:47
I couldnt find the limit – user63112 Feb 20 '13 at 15:33
-1: Your posts show no effort of your own on the problems. – JavaMan Feb 20 '13 at 18:42

Using the ratio test; calculate $$\lim_{n\to\infty}\left|\frac{\,\,\frac{(x-4)^{(n+1)^2}}{(n+1)!}\,\,}{\frac{(x-4)^{n^2}}{n!}}\right|=\lim_{n\to\infty}\left|\frac{n!}{(n+1)!}\cdot\frac{(x-4)^{(n+1)^2}}{(x-4)^{n^2}}\right|=\lim_{n\to\infty}\left|\frac{(x-4)^{2n+1}}{n+1}\right|.$$Now it is easy to check that for $3\leq x\leq5$, the series absolutely converges, as the limit above goes to $0$, and for $x\notin [3,5]$, the series diverges.
To specifically use the Root Test, notice that by Stirling's Approximation Formula, we have $$\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\frac{|x-4|^n\cdot e}{n}.$$ Now again we can see that $3\leq x\leq 5$ satisfies the condition for the series to absolutely converge; however it diverges for every $x\notin [3,5]$.
How can I show that $\lim_{x\rightarrow \infty } (x!)^{1/n} = 1$ ? – user63112 Feb 20 '13 at 16:04