Radius of Convergence for \sum_{n=2}^{\infty}z^n\log^2(n) in Complex Numbers

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
fauboca
Messages
157
Reaction score
0
[tex]\sum_{n=2}^{\infty}z^n\log^2(n), \ \text{where} \ z\in\mathbb{C}[/tex]

[tex]\sum_{n=2}^{\infty}z^n\log^2(n) = \sum_{n=0}^{\infty}z^{n+2}\log^2(n+2)[/tex]

By the ratio test,

[tex]\lim_{n\to\infty}\left|\frac{z^{n+3}\log^2(n+3)}{z^{n+2}\log^2(n+2)}\right|[/tex]

[tex]\lim_{n\to\infty}\left|z\left(\frac{\log(n+3)}{ \log (n+2)}\right)^2\right| = |z|[/tex]

if [itex]|z|<1[/itex], then the sum converges, and if [itex]|z|>1[/itex], then the sum diverges.

Does this mean that [itex]R=1[/itex]?
 
Last edited:
Physics news on Phys.org
LCKurtz said:
Yes, and there was no need to shift the indices.

Thanks.