The radius of convergence of $\sum_{n\ge 0}{\log(n!)x^n}$.

Question

I want the radius of convergence of the series $\sum_{n\ge 0}{\log(n!)x^n}$. Could I use the stirling formula $$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}?$$ Because then $$\log (n!)\sim_\infty \log\left(\middle(\frac{n}{e}\middle)^n\sqrt{2 \pi n}\right)$$

Then use a ratio test to compute the limit of $$\frac{\log(n+1)!}{\log(n!)}|x|$$

score 4 · Accepted Answer · answered Nov 16 '13 at 18:04

4

Note that $n!\leqslant n^n$ hence $1\leqslant\log(n!)\leqslant n\log n\leqslant n^2$ for every $n\geqslant3$.

The radii of convergence of $\sum\limits_nx^n$ and $\sum\limits_nn^2x^n$ are both $1$ hence the radius of convergence of $\sum\limits_n\log(n!)x^n$ is $____$.

If one insists on using the ratio test, one might note that $\log(n!)\sim n\log n$ hence $\log(n!)\sim\log((n+1)!)$, that is, $$ \frac{\log((n+1)!)}{\log(n!)}\to1, $$ thus, indeed, the radius of convergence of $\sum\limits_n\log(n!)x^n$ is $____$.

answered Nov 16 '13 at 18:04

Did

279,727

Thank you Did!! 1) The first inequalites are for $n\ge 3$ and the series starts at $n=0$, is this without influence? 2) What about the stirling formula and the equivalence I gave, is what i wrote correct? – palio Nov 16 '13 at 18:11
Yes this is without influence since the convergence of a series does not depend on its first terms. 2) What you wrote is correct but needs to be pushed on before being called a proof.

Did

Nov 16 '13 at 18:57

I may be a bit off-topic, but Did can you please elaborate on why knowing the radii of $\sum x^{n} $ and $ \sum n^{2}x^{n}$ helps ? I'm new to power series so my reasoning about it is not spot on yet. – Gabriel Romon Nov 16 '13 at 19:02

@GabrielR. This looks like a new MSE question to me. If you ask it (which I encourage you to do), please think hard about its formulation... – Did Nov 16 '13 at 21:30

I got it. You just used radius comparison since $x^n \leq ln(n!) x^{n} \leq n^{2}x^{n}$ – Gabriel Romon Nov 17 '13 at 11:28

score 2 · Answer 2 · answered Nov 16 '13 at 18:46

Because $$ \frac{\log((n+1)!)}{\log(n!)} = 1 + \frac{\log(n+1)}{\log(n!)}, $$ to prove that $\lim_{n\to\infty} \big( \frac{\log((n+1)!)}{\log(n!)} |x| \big) = |x|$ you only need to show that $\lim_{n\to\infty} \frac{\log(n+1)}{\log(n!)} = 0$.

This can be done, for instance, by observing that for $n\ge3$, we have $n+1 \le ((n+1)/2)^2$ and $$ n! > n(n-1)\cdots\bigg(n-\bigg\lceil \frac n2\bigg\rceil+1\bigg) > \bigg(\frac{n+1}2\bigg)^{\lceil n/2\rceil} \ge \bigg(\frac{n+1}2\bigg)^{n/2}, $$ and so $$ 0 \le \frac{\log(n+1)}{\log(n!)} < \frac{2\log((n+1)/2)}{\frac n2\log((n+1)/2)} = \frac 4n. $$ The squeeze theorem finishes it off.

The radius of convergence of $\sum_{n\ge 0}{\log(n!)x^n}$.

2 Answers2