Find integers a such that ∑ (n!)³ / (an)! converges

by RoRi

March 29, 2016

Find all positive integers $a$ such that the series

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

converges.

We use the ratio test,

$\begin{align*} \lim_{n \to \infty} \left( \frac{((n+1)!)^3}{(a(n+1))!} \cdot \frac{(an)!}{(n!)^3} \right) = \lim_{n \to \infty} \frac{(n+1)^3}{(an+1)(an+2)\cdots (an+a)}. \end{align*}$

Since the largest power of $n$ in the denominator is $a^n n^a$ , we have that this limit diverges for $a = 1, 2$ . If $a = 3$ then the limit is $\frac{1}{9}$ , and the limit is 0 if $a > 3$ . Thus, the series converges for all $a \geq 3$ .