# convergence properties of the series $\sum_{n=1}^\infty n^\alpha z^n$

when does the series $\sum_{n=1}^\infty n^\alpha z^n$ converges for $\alpha > 0$, I am tring to use $\lim_{n_\to \infty} \frac{|a_{n+1} |}{|a_{n}|}$ where $a_{n} = n^\alpha$

This gives me radius of convergence equal to 1, so the series converges for |z| <1 ?? I think this is wrong... do i need to consider $\alpha$ for the series to converge? how?

• Why do you think this is wrong? – Thomas Andrews Apr 12 '17 at 16:59
• Ratio test is inconclusive when the limit is $1$. Your computation shows that the series converges when $|z| < 1$ and diverges when $|z| > 1$, but tells nothing about the case $|z| = 1$. In your case, the situation is simple because the general term $n^{\alpha}z^n$ diverges when $\alpha > 0$ and $|z| = 1$. So you can safely conclude that the series converges exactly when $|z| < 1$. – Sangchul Lee Apr 12 '17 at 17:01

## 2 Answers

$S = \sum_{n=1}^\infty z^n = f(z) = \frac{z}{1-z}$ converges for $|z| < 1$. You may then differentiate w.r.t. z to obtain $$\frac {\rm d S}{\rm d z } = \sum_{n=1}^\infty n z^{n-1} = \sum_{m=0}^\infty (m+1) z^{m} = 1 + \sum_{m=1}^\infty (m+1) z^{m} = 1 + S + \sum_{m=1}^\infty m z^{m} = \frac {\rm d f(z)}{\rm d z }$$

from which you can deduce $\sum_{m=1}^\infty m z^{m}$ as a function of $z$. Continuing with this process gives you $\sum_{m=1}^\infty m^k z^{m}$ for all $k$. Now since $m^\alpha < m^k$ for all $m \geq 1$ and some $k$, you have a majorizing series which converges for $|z| < 1$. Hence you series converges for all $\alpha \geq 1$ as long as $|z| < 1$.

Hadamard's formula: The radius of convergence is given by $$\frac1R=\limsup_{n\to\infty}\bigl(n^\alpha\bigr)^{\tfrac1n}=\limsup_{n\to\infty}\Bigl(n^{\tfrac1n}\Bigr)^\alpha=1^\alpha.$$