Sum convergence

Question

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

Answer 1

Following your work:

$$\frac{a_{n+1}}{a_n}=\frac{(3n+4)!}{7^{n+1}(n+1)!(2n+3)!}\cdot\frac{7^nn!(2n+1)!}{(3n+1)!}=$$

$$=\frac17\frac{(3n+2)(3n+3)(3n+4)}{(n+1)(2n+2)(2n+3)}\xrightarrow[n\to\infty]{}\frac{27}{28}<1$$

Answer 2

$$\frac{(3n+4)!(2n+1)!}{(n+1)7(2n+3)!(3n+1)!} = \frac{(3n+4)(3n+3)(3n+2)}{(n+1)7(2n+3)(2n+2)} \to \frac{27}{28}<1 $$ hence the series converges.

I only used the definition of the factorial.