Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Help finding the radius of convergence?
Find the radius of convergence (R), and the convergence interval, for the following power series:
∞
Σ (2^n / n!)(x - 1)^n ; (a = 1)
n=0
So far I've gotten lim(n→∞) (2^(n+1)*((x-1)^(n+1)) / (n+1)!) * (n! / ((2^n)(x-1)^n)). I have no idea what to do after that.
2 Answers
- QCLv 710 years agoFavourite answer
lim(n→∞) (2^(n+1)*(|x-1|^(n+1)) / (n+1)!) * (n! / ((2^n)*|x-1|^n))
= lim(n→∞) (2^(n+1)/2^n) * (n!/(n+1)!) * (|x-1|^(n+1) / |x-1|^n)
= lim(n→∞) 2 * 1/(n+1) * |x-1|
= 0
Power series converges for all x, so radius of convergence is infinite.
-- Ματπmφm --
- cidyahLv 710 years ago
Apply the Ratio Test.
a(n+1)/a(n) =2^(n+1) n! /(n+1)! (x-1)^(n+1) / 2^n (x-1)^n
=2 (x-1) / (n+1)
As n approaches infinity, 2(x-1)/(n+1) approaches 0
Radius of convergence is 0
