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

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.

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