Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
Use Stirling's Approximation: $(n!) \sim c \sqrt{n} \left(\frac{n}{e}\right)^n$. –  JavaMan Jan 30 '12 at 19:28
1  
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
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Thanks Davide. Can you please explain why the first inequality, the fourth line of your align environment, follows? –  Carter Jan 30 '12 at 19:42
    
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
    
I recall that inequality now, thank you. –  Carter Jan 30 '12 at 19:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.