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January 23rd, 2016, 10:43 AM  #1 
Newbie Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0  Find the radius of convergence, and the interval of convergence, of the series.
Power Series: Find the radius of convergence, and the interval of convergence of the series: $\displaystyle \sum_{n=1}^{\infty }\frac{\sqrt{n}(2x+3)^{n}}{n^{2}+1}$ I have tried to solve this problem as following: $\displaystyle \lim_{n\rightarrow \infty }\left  \frac{a_{n+1}}{a_{n}} \right =\lim_{n\rightarrow \infty}\left  \frac{\sqrt{n+1}(2x+3)^{n+1}.(n^2+1)}{\sqrt{n}(2x+ 3)^n.((n+1)^2+1)} \right = \left  2x+3 \right  $ well, I have skipped some steps here because of too many codings. so $\displaystyle 2< x< 1$ And: $\displaystyle x=1, \sum_{n=1}^{\infty }\frac{\sqrt{n}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}}{n^2(1+\frac{1}{n^2})}=\frac{1}{n^{3/2}}$ Convergence, according to the root test $\displaystyle x=2, \sum_{n=1}^{\infty }\frac{\sqrt{n}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(1)^n}{n^2+1}=\sum_{n=1}^{\infty }\frac{n^{1/2}(1)^n}{n^2(1+\frac{1}{n^2})}=\frac{(1)^n}{n^{3/2}}$ Leibzig gives: 1: The term changes sign for each n 2: $\displaystyle a_{n+1}> a_{n}$ 3: $\displaystyle \lim_{n\rightarrow \infty } a_{n}=0$ But my fussy teacher says that everything is ok except my final answer and it should be written in a correct order (mathematically). So could you please look and tell me what is wrong? thank you. Last edited by skipjack; January 23rd, 2016 at 02:26 PM. 
January 23rd, 2016, 11:05 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,546 Thanks: 1259 
$2<x<1$ did you check the endpoints of the interval for convergence? 
January 23rd, 2016, 02:34 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,234 Thanks: 496 
At x = 2, it is absolutely convergent (same expression as x = 1), therefore it is convergent. Convergence at both ends gives interval $\displaystyle 2\leq x \leq1$. 

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convergence, find, interval, radius, series 
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