# Find the interval of convergence of the power series. sum_{n=0}^inftyfrac{x^{5n}}{n!}

Question
Series
Find the interval of convergence of the power series.
$$\sum_{n=0}^\infty\frac{x^{5n}}{n!}$$

2021-03-07
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
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 $$a_n=\frac{x^{5n}}{n!}$$
$$L=\lim_{n\to\infty}\left|\frac{\frac{x^{5n+5}}{(n+1)!}}{\frac{x^{5n}}{n!}}\right|$$
$$L=\lim_{n\to\infty}|\frac{n!}{x^{5n}}\frac{x^{5n+5}}{(n+1)!}|$$
$$L=\lim_{n\to\infty}|\frac{n!}{x^{5n}}\frac{x^{5n}\times x^5}{(n+1)(n)!}|$$
$$L=\lim_{n\to\infty}|\frac{x^5}{(n+1)}|$$
$$L=|x^5|\lim_{n\to\infty}\frac{1}{(n+1)}$$
$$L=|x^5|\times0$$
$$L=0<1$$</span>
Here the limit is less than 1, and independent of the value of x.
Hence the given power series is convergent for all $$x\in(-\infty,\infty)$$

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