I am looking for an explanation that is easily explainable to a Calc. II student and makes use of the usual convergence tests.
Determine the convergence/divergence of $$\sum_{n=1}^{\infty}\dfrac{(n!)^2}{(n^2)!}$$
By the ratio test, we have $$\begin{align}\dfrac{[(n+1)!]^2}{[(n+1)^2]!} \cdot \dfrac{(n^2)!}{(n!)^2}&=\dfrac{(n+1)^2}{(n+1)^2[(n+1)^2-1][(n+1)^2-2]\cdots(n^2+1)} \\ &= \dfrac{1}{[(n+1)^2-1][(n+1)^2-2]\cdots(n^2+1)} \end{align}$$ and taking the limit as $n \to \infty$, we obtain $0$. Hence, the series is absolutely convergent, and thus convergent.
Is this correct? If so, is there perhaps an easier way to approach this problem?
Since the series is with positive terms we can apply ratio test and since the limit of the ratio is $0$ yes it is convergent (we don’t need to refer to absolute convergence).
As an alternative we could use Stirling but I think that ratio test is really a nice way in this case.