What is the convergence radius for the series #sum_(r=0)^oo((x+1)/3)^r# ?

1 Answer
Nov 20, 2016

Answer:

#abs(x+1)<3#

Explanation:

You can write this series as

#sum_(r=0)^oo((x+1)/3)^r# Calling now #y = (x+1)/3# we have equivalently

#lim_(n->oo)sum_(r=0)^ny^r=lim_(n->oo)(y^n-1)/(y-1)#.

We are using the fact

#1+y+y^2+cdots+y^n=(y^(n+1)-1)/(y-1)#

Now, if #absy < 1# we have #lim_(n->oo)y^(n+1)=0# so assuming #absy < 1#

#lim_(n->oo)sum_(r=0)^ny^r = 1/(1-y) = 1/(1-(x+1)/3) = 3/(2-x)# this of course for #abs(x+1) < 3#