• May 7th 2009, 10:08 AM
fattydq
I'm trying to find the radius of convergence of the problem I've attached (the top one in the attached pic) I know you have to use the ratio test somehow, but then it just seems like that leaves me with a very complex number that I can't really figure out. Help a brother out?
• May 7th 2009, 10:27 AM
Calculus26
Cn = 1/(n^2+1)

C(n+1) = 1/[(n+1)^2+1] = 1/(n^2 +2n +2)

lim|(x-5)^(n+1)C(n+1)/[(x-5)^nCn ]|< 1

|x-5| lim (n^2+1)/(n^2 + 2n +1) < 1

lim (n^2+1)/(n^2 + 2n +1) = 1 Use L'hopital's Rule on the corresponding continuous function if you need or note the lim is just the limit of the highest powered terms.

so the radius of convergence is 1

|x-5| < 1

We have convergence for 4 < x < 6

Check x= 4 and x =6 in the original series to determine the convergence issue at the endpoints
• May 26th 2009, 08:35 PM
Showcase_22
Quote:

so the radius of convergence is 1
To find a radius of convergence, shouldn't we be looking for some $a \in \mathbb{R}$ s.t when $|x| the series converges?

Therefore $|x-5|<1 \Rightarrow 4.

Hence the radius of convergence would be 4?
• May 27th 2009, 10:49 AM
Moo
Quote:

Originally Posted by Showcase_22
To find a radius of convergence, shouldn't we be looking for some $a \in \mathbb{R}$ s.t when $|x| the series converges?

Therefore $|x-5|<1 \Rightarrow 4.

Hence the radius of convergence would be 4?

But the series wouldn't converge for $x<4$ (Surprised)

You can let $z=x-5$ and then say that the radius of convergence of the new series is 1.

And you may actually be able to say that it converges on the disk of center 5, and radius 1.