Find the interval of convergence of the power series. sum_{n=0}^inftyfrac{(3x)^n}{(2n)!}

Question
Series
asked 2021-02-03
Find the interval of convergence of the power series.
\(\sum_{n=0}^\infty\frac{(3x)^n}{(2n)!}\)

Answers (1)

2021-02-04
Interval of convergence using ratio test: If
\(\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=L\)
If L
If L>1 then the series is absolutely divergent
Given that
\(a_n=\frac{(3x)^n}{(2n)!}\)
So \(a_{n+1}=\frac{(3x)^{n+1}}{(2n+2)!}\)
Now evaluate \(|\frac{a_{n+1}}{a_n}|\), so
\(|\frac{a_{n+1}}{a_n}|=|\frac{\frac{(3x)^{n+1}}{(2n+2)!}}{\frac{(3x)^n}{(2n)!}}|\)
\(=|\frac{(3x)^{n+1}}{(2n+2)!}\frac{(2n)!}{(3x)^n}|\)
\(=|\frac{(3x)}{(2n+2)(2n+1)}|\)
Now
\(L=\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|\)
\(=\lim_{n\to\infty}|\frac{(3x)}{(2n+2)(2n+1)}|\)
\(=|3x|\lim_{n\to\infty}|\frac{1}{n^2(2+\frac{2}{n})(2+\frac{1}{n})}|\)
\(L=0,\)
Here L = 0
Hence interval of convergence is \((-\infty,\infty)\).
0

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