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

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

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 $$(-\infty,\infty)$$.

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