<s>[INST] Sum convergence I want to check this sum: $$\sum\limits_{n=1}^\infty (\frac{(3n+1)!}{n!(2n+1)!}*7^{-n})$$
I think, the easiest way is to use the ratio test:
$$\frac{\frac{(3n+4)!7^{-(n+1)}}{(n+1)!(2n+1)!}}{\frac{(3n+1)!}{n!(2n+1)!}7^{-n}} =\frac{(3n+4)!7^{-(n+1)}n!(2n+1)!}{(n+1)!(2n+3)!(3n+1)!7^{-n}}=\frac{(3n+4)!(2n+1)!}{(n+1)7(2n+3)!(3n+1)!}$$
But how should I proceed?
Thank you for your help
queenD
 [/INST] Following your work:
$$\frac{a_{n+1}}{a_n}=\frac{(3n+4)!}{7^{n+1}(n+1)!(2n+3)!}\cdot\frac{7^nn!(2n+1)!}{(3n+1)!}=$$
$$=\frac17\frac{(3n+2)(3n+3)(3n+4)}{(n+1)(2n+2)(2n+3)}\xrightarrow[n\to\infty]{}\frac{27}{28}<1$$
</s>