My Math Forum interval / radius of convergence

 Calculus Calculus Math Forum

 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 )((2x-1)^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!

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Johanovegas Calculus 2 January 23rd, 2016 01:34 PM leo255 Calculus 2 November 8th, 2014 05:11 PM Singularity Calculus 6 April 20th, 2014 12:17 PM hbarnes Calculus 2 July 19th, 2013 07:00 AM Makino Calculus 1 June 24th, 2009 08:23 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top