Find the radius of convergence, R, of the series. ∑n=1∞(2x+9)nn2 Find the interval, I, of convergence of the series.

Question

Find the radius of convergence, R, of the series.  ∑n=1∞(2x+9)nn2  Find the interval, I, of convergence of the series.

doplovif · Accepted Answer

Given series is ∑n=1∞(2x+9)nn2 an=(2x+9)nn2 We will use Ratio Test to find radius of convergence of the series. limn→∞|an+1an|=limn→∞|(2x+9)n+1(n+1)2⋅n2(2x+9)n|=limn→∞|(2x+9)n2(2x+9)n|=|2x+9| By Ratio Test , given series converges for |2x+9|&amp;lt;1 2|x−(92)|&amp;lt;1 |x−(92)|&amp;lt;12 Therefore, radius of convergence is R=12 Now, |2x+9|&amp;lt;1 −1&amp;lt;2x+9&amp;lt;1 −1−9&amp;lt;2x&amp;lt;1−9 −10&amp;lt;2x&amp;lt;−8 −5&amp;lt;x&amp;lt;−4 When x=5, series become ∑n=1∞(2(−5)+9)nn2=∑n=1∞(−1)nn2 Which is convergent by Alternating series test ( because sequence 1n2 is decreasing and converging to 0 ) When x=−4, series become ∑n=1∞(2(−4)+9)nn2=∑n=1∞(−1)nn2=∑n=1∞1n2 Which is also convergent by p-series test. Hence, interval of convergence is [−5,−4] And: Radius of convergence is R=12 Interval of convergence is [−5,−4]

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.

Answered question

Answer & Explanation

New Questions in Calculus 2