Question

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.

Series
ANSWERED
asked 2021-03-18
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.

Answers (1)

2021-03-19

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 \([-5,-4]\)

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