# Thread: Finding a convergence radius

1. ## Finding a convergence radius

Find the convergence radius of $\sum_{n=0}^{\infty} n! \pi^{\frac{n}{2}}x^{n!}$

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My attempt:

Define $a_n:= n! \pi^{\frac{n}{2}}x^{n!}$

Therefore $\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{(n+1)!\pi^{\frac{n+1}{2}}x^{(n+1)!}}{n! \pi^{\frac{n}{2}} x^{n!}} \right|$

$=\left| (n+1) \pi^{\frac{1}{2}} x^{n! n} \right|$

If $|x|<1$ then $R=\infty$.
If $|x|\geq 1$ then $R=0$

Is this right?

2. Well I think you should have that $|\frac{a_{n+1}}{a_n}|=c_n=|(n+1)\sqrt{\pi}x^{n+1}|$.

So you need $\lim c_n<1\Longrightarrow \lim_n|nx^n|<\frac{1}{\sqrt{\pi}}$. Which is only true if $|x|<1$. Thus the radius of convergence is $R=1$.