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.

Set $k = rm$, you get the expression as 
$$\left( \frac{(n!)^r}{(2n)!} \right)^\frac{1}{m}$$
Use ratio test after that, you should get
$$\left| \frac{(n+1)^r}{(2n+2)(2n+1)}\right| < 1$$
