# Find the radius of convergence, R, of the series. sum_{n=1}^inftyfrac{(2x+9)^n}{n^2} Find the interval, I, of convergence of the series.

Find the radius of convergence, R, of the series.
$$\sum_{n=1}^\infty\frac{(2x+9)^n}{n^2}$$
Find the interval, I, of convergence of the series.

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doplovif

Given series is
$$\sum_{n=1}^\infty\frac{(2x+9)^n}{n^2}$$
$$a_n=\frac{(2x+9)^n}{n^2}$$
We will use Ratio Test to find radius of convergence of the series.
$$\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to\infty}|\frac{(2x+9)^{n+1}}{(n+1)^2}\cdot\frac{n^2}{(2x+9)^n}|=\lim_{n\to\infty}|(2x+9)\frac{n^2}{(2x+9)^n}|=|2x+9|$$
By Ratio Test , given series converges for
$$|2x+9|<1$$
$$2|x-(\frac{9}{2})|<1$$
$$|x-(\frac{9}{2})|<\frac{1}{2}$$
Therefore, radius of convergence is $$R=\frac{1}{2}$$
Now,
$$|2x+9|<1$$
$$-1<2x+9<1$$
$$-1-9<2x<1-9$$
$$-10<2x<-8$$
$$-5<x<-4$$
When $$x=5$$, series become
$$\sum_{n=1}^\infty\frac{(2(-5)+9)^n}{n^2}=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}$$
Which is convergent by Alternating series test ( because sequence $$\frac{1}{n^2}$$ is decreasing and converging to 0 )
When $$x=-4$$, series become
$$\sum_{n=1}^\infty\frac{(2(-4)+9)^n}{n^2}=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=\sum_{n=1}^\infty\frac{1}{n^2}$$
Which is also convergent by p-series test.
Hence, interval of convergence is $$[-5,-4]$$
And:
Radius of convergence is $$R=\frac{1}{2}$$
Interval of convergence is $$[-5,-4]$$