Radius of convergence of this power series

Question

We're given the power series $$ \sum_1^\inf \frac{j!}{j^j}z^j$$

and are asked to find radius of convergence R. I know the formula $R=1/\limsup(a_n ^{1/n})$, which leads me to compute $\lim \frac{j!^{1/j}}{j}$, and then I'm stuck.

The solution manual calculates R by $1/\lim|\frac{a_{j+1}}{a_j}|$, but I can't figure out the motivation for that formula.

See for example this question. – Daniel Fischer Oct 23 '15 at 14:35 — Daniel Fischer, Oct 23 '15 at 14:35

score 1 · Answer 1 · answered Oct 23 '15 at 14:30

1

Use Stirling's formula $$n! \sim {n^n e^{-n}\over\sqrt{2\pi n}}.$$

answered Oct 23 '15 at 14:30

ncmathsadist

49,383

score 1 · Answer 2 · answered Oct 23 '15 at 14:34

1

The ratio test gives that, if $\{a_n\}$ is a positive sequence and $\frac{a_{n+1}}{a_n}\to L$, then $$ \lim_{n\to +\infty}\sqrt[n]{a_n} = L, $$ too. You may also notice that by Lagrange's inversion your series is related with the inverse function of $x\mapsto x\cdot e^x$, i.e. the Lambert $W$-function.

answered Oct 23 '15 at 14:34

Jack D'Aurizio

353,855

1

Thanks! now I remember reading about that result in an earlier chapter. – Anton Fahlgren Oct 23 '15 at 14:36

Radius of convergence of this power series

2 Answers2