May 1st, 2018, 06:13 AM  #1 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0  interval / radius of convergence
Any guidance will be appreciated : find the interval of convergence u_n=((2^n (n+2))/n^2 )((2x1)^n) 
May 1st, 2018, 10:20 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
A pretty standard method for power series is to use the "ratio test". Here $\displaystyle u_n= \frac{2^n(n+2)}{n^2}(2x 1)^n$ and $\displaystyle u_{n+1}= \frac{2^{n+1}((n+1)+ 2)}{(n+1)^2}(2x 1)^{n+1}$$\displaystyle = \frac{2(2^n)(n+ 3)}{(n+1)^2}(2x 1)^{n+1}$ So $\displaystyle \frac{u_{n+1}}{u_n}=$$\displaystyle \frac{2^{n+1}}{2^n}\frac{n+3}{n+1}\frac{(n+1)^2}{n ^2}\frac{(2x 1)^{n+1}}{(2x 1)^n}$$\displaystyle = \frac{(2^{n+1}}{2^n}\frac{n+3}{n+1}\frac{(n+1)^2}{ n^2}(2x 1)^1$ Take the limit as n goes to infinity, $\displaystyle \frac{2^{n+1}}{2^n}= 2$, $\displaystyle \frac{n+2}{n+1}$ goes to 1, $\displaystyle \frac{n+1)^2}{n^2}= \frac{n^2+ 2n+ 1}{n^2}= 1+ \frac{2}{n}+ \frac{1}{n^2}$ goes t0 1 so this all goes to $\displaystyle 2(2x 1)$. The series converges, absolutely, for the absolute value of that fraction less than 1: $\displaystyle 4x 2< 1$ so 1< 4x 2< 1, 1< 4x< 3, 1/4< x< 3/4. 
May 1st, 2018, 09:04 PM  #3 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
Thanks for the help! 

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convergence, interval, radius 
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