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$$\sum_{n=1}^{\infty} \frac{(x-4)^{n^{2}}}{n!}$$ 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
up vote 5 down vote accepted

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 $x\notin [3,5]$.

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How can i solve this problem by the root test? – user63112 Feb 20 '13 at 15:39
@user63112: If you like the answer; please be sure to accept it (clicking the check mark beneath the up/down arrows to the left). If you still have other questions, feel free to ask – Clayton Feb 20 '13 at 15:57
@clayton the nth root of n factorial is approximate to \frac{n}{e} for n big – Dominic Michaelis Feb 20 '13 at 15:58
How can I show that $\lim_{x\rightarrow \infty } (x!)^{1/n} = 1$ ? – user63112 Feb 20 '13 at 16:04
@user63112: I have updated my solution to include the Root Test. – Clayton Feb 20 '13 at 18:01

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