Solving for Radius of Convergence: 1/(1+x^2) and arctan (x)

[rootX]

[SOLVED] Radius of Convergence

Homework Statement

1/(1+x^2) = sum ( (-1)^k*x^(2k) ; 0 ; inf) - A

integrating

arctan (x) = sum ((-1)^k * x^(2k+1) / (2k+1) ; 0; inf) B

I know A has radius of converge of 1, and I calculated B to be 2.

My assignment solution says "Similarly, the series for 1/(1+x^2) has R = 1 and integrating does not affect this. so R for atan (x) series is 1"

Obviously, they are wrong . Right?

Thanks.

 

[nicksauce]

Your notation ;0;inf)-A and ;0;inf)B is confusing. Can you clarify?

 

[rootX]

Sorry for the confusion.
It says sum of "(-1)^k * x^(2k+1) / (2k+1) "
k is from 0 to inf

[they are series]

 

[HallsofIvy]

Homework Statement

1/(1+x^2) = sum ( (-1)^k*x^(2k) ; 0 ; inf) - A

integrating

arctan (x) = sum ((-1)^k * x^(2k+1) / (2k+1) ; 0; inf) B

I know A has radius of converge of 1, and I calculated B to be 2.

My assignment solution says "Similarly, the series for 1/(1+x^2) has R = 1 and integrating does not affect this. so R for atan (x) series is 1"

Obviously, they are wrong . Right?

Thanks.

Well, there is one other possiblity!

Unfortunately, since you don't say HOW you got 2 as the radius of convergence for B, there isn't a whole lot I can say.

Using the ratio test,
\frac{|x^{2k+3}|}{2k+3}\frac{2k+1}{|x^{2k+1}|}= \frac{2k+1}{2k+3}|x|^2< 1
gives |x|< 1. Radius of convergence 1.

Did you forget the "2" on 2k+ 3?

 

[rootX]

Thanks a lot,

I saw 2 in there (2^2k+1), and made it
abs(x^2)/4 <1
without going through all the steps.