Consider the given power series:

\(\sum a_n=\sum_{n=1}^\infty\frac{x^{5n}}{n!}\)

Here the objective is to find the interval of x for which the given power series is convergent.

According to the ratio test

\(L=\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|\)

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

\(L=\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|\)

Substitute \(a_{n+1}=\frac{x^{5n+5}}{(n+1)!}\) and \(a_n=\frac{x^{5n}}{n!}\)

\(L=\lim_{n\to\infty}\left|\frac{\frac{x^{5n+5}}{(n+1)!}}{\frac{x^{5n}}{n!}}\right|\)

\(L=\lim_{n\to\infty}|\frac{n!}{x^{5n}}\frac{x^{5n+5}}{(n+1)!}|\)

\(L=\lim_{n\to\infty}|\frac{n!}{x^{5n}}\frac{x^{5n}\times x^5}{(n+1)(n)!}|\)

\(L=\lim_{n\to\infty}|\frac{x^5}{(n+1)}|\)

\(L=|x^5|\lim_{n\to\infty}\frac{1}{(n+1)}\)

\(L=|x^5|\times0\)

\(L=0<1\)

Here the limit is less than 1, and independent of the value of x.

Hence the given power series is convergent for all \(x\in(-\infty,\infty)\)