- #1

broegger

- 257

- 0

[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?