# Why are $\limsup \sqrt[n]{1/n!}=0$ and $\limsup\sqrt[n]{n!}=\infty$?

I was looking at the power series $\sum\frac{z^n}{n!}$ and $\sum n!z^n$, and wanted to compute their radii of convergence.

For the first, $\limsup \sqrt[n]{1/n!})=0$, and for the second $\limsup \sqrt[n]{n!}=\infty$, so the radii are $\infty$ and $0$ respectively.

I'm curious, without resorting to Wolfram or Mathematica or some similar program, how could one justify by hand these limits? Intuitively they make sense that they are what they are, since the factorial should grow more quickly than the power of $1/n$ can handle. I suppose would explain the other since the terms are just reciprocals. Thanks!

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Use Stirling's Approximation: $(n!) \sim c \sqrt{n} \left(\frac{n}{e}\right)^n$. –  JavaMan Jan 30 '12 at 19:28
Also, notice that $\sqrt[n]{n!} \to \infty$ implies that $\frac{1}{\sqrt[n]{n!}} \to 0$, so your question is really about how $\lim_{n \to \infty} \sqrt[n]{n!} = \infty$. –  JavaMan Jan 30 '12 at 19:33
We just have to show that $\lim_{n\to\infty}\frac 1n\sum_{k=1}^n\ln k=+\infty$. We have \begin{align*} \frac 1n\sum_{k=1}^n\ln k&=\frac 1n\sum_{k=1}^n\left(\ln \frac kn+\ln n\right)\\ &=\ln n+\frac 1n\sum_{k=1}^n\ln\left(1+\frac{k-n}n\right)\\ &=\ln n+\frac 1n\sum_{j=0}^{n-1}\ln\left(1-\frac jn\right)\\ &\geq \ln n +\frac 1n\sum_{j=0}^{n-1}-\frac jn-\frac{j^2}{n^2}\\ &\geq \ln n+\frac 1nn\left(-\frac{n-1}n-\frac{(n-1)^2}{n^2}\right)\\ &=\ln n-1+\frac 1n-\frac{(n-1)^2}{n^2}\\ &\geq \ln n-2, \end{align*} which gives the result.
It's $\ln(1+t)\geq t-\frac{t^2}2$, which can be shown for example using Taylor's formula. –  Davide Giraudo Jan 30 '12 at 19:46