At what $n$ does $\sum_{k=0}^\infty{(n\ k)!\over (k!)^3}$ diverge?

Question

Consider $$\sum_{k=0}^\infty{(n\ k)!\over (k!)^3}$$

where $0<n$.

For what $x$ where $0<x\leq n$ does the sum diverge?

Answer 1

obviously converges for $n=1$, $n=2$ is also easy, because $\frac{5^k}{k!}$ converges (ratio test), and since $(2k)! \leq (k!)^2 5^k$ for $k$ sufficiently large (induction*) we get what we want. for $n \geq 3$ it diverges, because $(3k)! \geq (k!)^3$ \ induction goes as follows - assume it's true for $k$, then for $k+1$ you need to show that $$ (2k+1)(2k+2) \leq 5(k+1)^2$$ which is true for k sufficiently large, because on the LHS you have $4k^2$ + something of lower order and on the RHS you have $5k^2$