Find the interval of convergence of the power series. ∑n=0∞x5nn!

Consider the given power series:
$\sum a_n=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{5n}}{n!}$
Here the objective is to find the interval of x for which the given power series is convergent.
According to the ratio test
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$
If $L<1$ then the series converges absolutely
If $L>1$ then the series is divergent
Here $a_n=\frac{x^{5n}}{n!}$ Replace $n\to n+1$
$a_{n+1}=\frac{x^{5n+5}}{(n+1)!}$
Use ratio test for convergence
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$
Substitute $a_{n+1}=\frac{x^{5n+5}}{(n+1)!}$ and $a_n=\frac{x^{5n}}{n!}$
$L=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\frac{x^{5n+5}}{(n+1)!}}{\frac{x^{5n}}{n!}}\right|$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{n!}{x^{5n}}\frac{x^{5n+5}}{(n+1)!}|$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{n!}{x^{5n}}\frac{x^{5n}\times x^5}{(n+1)(n)!}|$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{x^5}{(n+1)}|$
$L=|x^5|\underset{n\to \mathrm{\infty }}{lim}\frac{1}{(n+1)}$
$L=|x^5|\times 0$
$L=0<1$
Here the limit is less than 1, and independent of the value of x.
Hence the given power series is convergent for all $x\in (-\mathrm{\infty },\mathrm{\infty })$

Answer is given below (on video)