Radius Of Convergence of the series

Question

I want to calculate the radius of convergence of the series $\sum^{\infty}_{n=0} x^{n!}$.

I've tried the ratio test to no avail, obtaining:

$\frac{|x^{(n+1)!}|}{|x^{n!}|}= |x^{(n+1)!-n!}| = |x^{n!n}|$

I'm sure to go about applying the root test. If anyone could point me in the right direction that would be great.

Answer 1

Another option to find the radius of convergence is to note that for $|x|<1$, we have that $$\sum_{n=0}^\infty |x|^{n!} < \sum_{n=0}^\infty |x|^n$$ and so the series converges absolutely by the comparison test. (The geometric series contains all of the terms in the sum with factorials as exponents, and then some.)

On the other hand, we see that the series clearly diverges for $x=1$ (it is the sum of infinitely many terms all equal to $1$) We conclude that the radius of convergence must be $1$.

Answer 2

Well root test tells you to look at $\lim_{n\rightarrow \infty}\sqrt[n]{x^{n!}}=\lim_{n\rightarrow \infty}x^{(n-1)!}$. The conclusions method is the same as the conclusions of the ratio test.