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

Question
Series
asked 2021-03-06
Find the interval of convergence of the power series.
\(\sum_{n=0}^\infty\frac{x^{5n}}{n!}\)

Answers (1)

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 \(x\in(-\infty,\infty)\)
0

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