1. May 30, 2009

garyljc

Hey ,
I was wondering if anyone could help me out with this question regarding calculating the radius of convergence of the infinity series of (1/n!)x^(n!)

This is my work

First we consider when abs(x) < 1
then we have 0 <= abs(x^n!) <= abs(x^n)
so we know that the series converges whenever abs(x) < 1 , so by the comparison test, we conclude that the series converges for abs(x) < 1

Now consider abs(x) > 1
we have 0<= abs (x) <= abs(x^n!)
so the series by comparison test diverges for abs(x) > 1

After that , I consider what happen at x=1 and x=-1 , it turned out that they converges as well .
So my radius of convergence is [-1,1] and the series converges for abs(x) <= 1

is this approach correct ? because i thought that for question like this , ratio test is not a good approach.

2. May 30, 2009

Staff: Mentor

I don't see anything wrong, except that your radius of convergence, R, is 1. The interval of convergence is [-1, 1], and the radius of this interval is 1.

3. May 31, 2009

garyljc

Alright thanks a lot =D .