I shall provide the whole problem. To prove is the absolute convergence of
$ \sum_{i=0}^{\infty}(-1)^{3n} \dfrac{x^{4n}}{(5n)!}$
I decide to solve this problem using the Quotient Criteria and eventually end up with
$ \dfrac{\dfrac{(-1)^{n+1} x^{4(n+1)}}{(5(n+1))!}}{\dfrac{(-1)^{n} x^{4n}}{(5n)!}} = \dfrac{(-1)^{n+1} x^{4(n+1)} (5n)!}{(5(n+1))! (-1)^{n} x^{4n}} = - \dfrac{x^4 (5n)!}{(5(n+1))!}$
Now I just need to prove that $-\dfrac{x^4 (5n)!}{(5(n+1))!}$ converges, which is where I get stuck. It's obvious to me that it converges to $0$ since the bottom part of the equasion grows much faster than the top part.
How do I work with the $(5(n+1))?$
Did I make a mistake?
$-\dfrac{x^4 (5n)!}{(5(n+1))!}$ = $-\dfrac{x^4 (5n)!}{(5n+5)!}$ = $-\dfrac{x^4 (5n)!}{(5n+5)...(5n+1)(5n)!} $ = $-\dfrac{x^4}{(5n+5)...(5n+1)}$
$\lim_{n \to +\infty} \dfrac{x^4}{(5n+5)...(5n+1)} = 0$