Find real c such that ∑ (n!)^c / (3n)! converges

by RoRi

March 29, 2016

Find all $c \in \mathbb{R}$ such that the following series is convergent:

$\sum_{n=1}^{\infty} \frac{(n!)^c}{(3n)!}.$

Consider the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{((n+1)!)^c}{(3(n+1))!} \cdot \frac{(3n)!}{(n!)^c} \right) \\[9pt] &= \lim_{n \to \infty} \left( \left( \frac{(n+1)!}{n!}\right)^c \cdot \frac{1}{(3n+1)(3n+2)(3n+3)} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(n+1)^c}{(3n+1)(3n+2)(3n+3)}. \end{align*}$

This limit is 0 if $c < 3$ and is $\frac{1}{9}$ if $c = 3$ . If $c > 3$ then the limit diverges. Hence, by the ratio test we have that the series converges for $c \leq 3$ and diverges for $c > 3$ .