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

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SUMMARY

The radius of convergence for the series \(\sum_{n=2}^{\infty}z^n\log^2(n)\) in complex numbers is determined using the ratio test. The analysis shows that \(\lim_{n\to\infty}\left|z\left(\frac{\log(n+3)}{\log(n+2)}\right)^2\right| = |z|\). The series converges for \(|z|<1\) and diverges for \(|z|>1\), confirming that the radius of convergence \(R\) is indeed 1. There was no necessity to shift the indices in the evaluation process.

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[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:
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Yes, and there was no need to shift the indices.
 
LCKurtz said:
Yes, and there was no need to shift the indices.

Thanks.
 

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