- #1
broegger
- 257
- 0
I am given this series:
First I have to find the radius of convergence; I find R = 1. Then I have to show that the series is convergent, but not absolutely convergent, for z = -1, i.e. that the series
is convergent, while
is not. There is a hint that says [tex]\frac{2n}{n^2+1}\leq\frac1{n}[/tex]. How can I do this, I know only of the most basic convergence tests?
[tex]\sum_{n=1}^\infty\frac{2n}{n^2+1}z^n.[/tex]
First I have to find the radius of convergence; I find R = 1. Then I have to show that the series is convergent, but not absolutely convergent, for z = -1, i.e. that the series
[tex]\sum_{n=1}^\infty(-1)^n\frac{2n}{n^2+1}[/tex]
is convergent, while
[tex]\sum_{n=1}^\infty\frac{2n}{n^2+1}[/tex]
is not. There is a hint that says [tex]\frac{2n}{n^2+1}\leq\frac1{n}[/tex]. How can I do this, I know only of the most basic convergence tests?