text: # 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.
  
  • Questions are typically answered in as fast as 30 minutes
  
  ### Plainmath recommends
  
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
  
  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 $$[-5,-4]$$
explanation: The document contains a math problem that asks for the radius of convergence and the interval of convergence of a given series. The text also provides a detailed solution to the problem, including the steps to determine the radius and interval of convergence.
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distilabel_metadata.role: system
distilabel_metadata.content: # 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.
  
  • Questions are typically answered in as fast as 30 minutes
  
  ### Plainmath recommends
  
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
  
  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 $$[-5,-4]$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem that asks for the radius of convergence and the interval of convergence of a given series. The text also provides a detailed solution to the problem, including the steps to determine the radius and interval of convergence.
  
  Label: question-and-answer