calculus - Convergence of $sum^infty_{n=1}frac {sqrt[m]{n!}}{sqrt[k]{(2n)!}}$

Does the following series converges ? $$\displaystyle\sum^\infty_{n=1}\frac {\sqrt[m]{n!}}{\sqrt[k]{(2n)!}} \ \text{for} \ \ k,m\in \mathbb N$$

I tried the ratio test:

$ \displaystyle\lim_{n\to\infty}\frac {\sqrt[m]{(n+1)!}}{\sqrt[k]{(2n+2)!}}\cdot \frac {\sqrt[k]{(2n)!}}{\sqrt[m]{n!}} = ... =\lim_{n\to\infty}\frac {(n+1)^{\large\frac{k-m}{mk}}}{(2(2n+1))^{\large\frac1 k}}$

Now I should check for cases with $m,k$ where the numerator is larger than the denominator and vice versa and when they're equal but it doesn't seem right...

Note: I can't use integration or Stirling approximation, nor Taylor.

Answer

If you look at the ratio you have and rewrite it slightly, you get

$$n^{\frac{1}{m} - \frac{2}{k}}\frac{\left(1+\frac{1}{n}\right)^{(k-m)/(mk)}}{2^{2/k}\left(1+\frac{1}{2n}\right)^{1/k}}.$$

The fraction converges to

$$\frac{1}{2^{2/k}} < 1,$$

so it depends on the behaviour of $n^{1/m - 2/k}$. If $\frac{1}{m} > \frac{2}{k}$, the quotient tends to $+\infty$, if $\frac{1}{m} < \frac{2}{k}$, it tends to $0$, and in the case of equality, it tends to $2^{-2/k} \in (0,1)$.

So by the ratio test, the series converges if and only if $k \leqslant 2m$.