- #1
garyljc
- 103
- 0
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.
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.
