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

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.

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