May 1st, 2018, 07:13 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, 11:20 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, 10:04 PM  #3 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
Thanks for the help! 

Tags 
convergence, interval, radius 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Find the radius of convergence, and the interval of convergence, of the series.  Johanovegas  Calculus  2  January 23rd, 2016 02:34 PM 
Did I do this interval/radius of convergence problem correctly?  leo255  Calculus  2  November 8th, 2014 06:11 PM 
Finding radius and interval of convergence  Singularity  Calculus  6  April 20th, 2014 01:17 PM 
Find the radius of convergence and the interval of conv...  hbarnes  Calculus  2  July 19th, 2013 08:00 AM 
Radius and interval of convergence  Makino  Calculus  1  June 24th, 2009 09:23 AM 