Given the sequence #a_1=sqrt(y),a_2=sqrt(y+sqrt(y)), a_3 = sqrt(y+sqrt(y+sqrt(y))), cdots# determine the convergence radius of #sum_(k=1)^oo a_k x^k# ?

Question

1549 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-08T20:39:50

If #y = 0# then the radius is infinite. Otherwise it is #1#.

Assuming we want to deal with Real numbers only, we require #y >= 0#

If #y=0# then the radius of convergence is infinite.

Otherwise, #y > 0#

Note that the sequence #a_1, a_2, a_3,...# is strictly monotonic increasing.

It does have a finite fixed point towards which it converges:

Let #t = sqrt(y+sqrt(y+sqrt(y+sqrt(y+sqrt(y+...)))))#

Then:

#t^2-t-y = 0#

So using the quadratic formula:

#t = (1+-sqrt(1+4y))/2#

and since #t >= 0# we must have:

#t = 1/2+sqrt(1+4y)/2 = 1/2+sqrt(y+1/4)#

This is a fixed point of the function #f(t) = sqrt(y+t)#

In particular, if #x >= 0# then

#sqrt(y) sum_(k=1)^oo x^k <= sum_(k=1)^oo a_k x^k <= (1/2+sqrt(y+1/4)) sum_(k=1)^oo x^k#

So #sum_(k=1)^oo a_k x^k# is absolutely convergent if and only if #sum_(k=1)^oo x^k# is absolutely convergent, which is if and only if #abs(x) < 1#.

Hence the radius of convergence is #1#.