What happens at the boundary of the disk of convergence?

Asked Sep 03 '17 at 15:12

Active Sep 03 '17 at 15:19

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I got stuck in solving a question of power series in complex analysis in Conway's exercise. Which is as follows $:$

Find the radius of convergence of the power series $:$

$$\sum_{n=1}^{\infty} \frac {(-1)^n} {n} z^{n(n+1)}.$$

What happens at the boundary of the disk of convergence?

I found that the radius of convergence to be $1$ and hence the disk of convergence is $|z|=1$. Now I found at $z=1,-1$ the series is convergent by Leibnitz test. Now how can I check the point of convergence (if any) in the disk of convergence other than $1$ and $-1$. I tried with Dirchilet's test. But I couldn't actually figure out how is $$\sum_{k=1}^{n} z^{k(k+1)}$$ look like? Is it bounded for arbitrary $n$ for all $z$ on $\partial B(0;1)$ except at $1$ and $-1$?

Please give me some suggestions.

Thank you in advance.

edited Jun 12 '20 at 10:38

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asked Sep 03 '17 at 15:12

Arnab Chattopadhyay.

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It's not hard to see that it also converges when $z=\pm i$. – José Carlos Santos Sep 03 '17 at 15:55
Yes.It's correct.But for all other points? – Arnab Chattopadhyay. Sep 03 '17 at 16:07
Use partial summation to make $\sum_{m \le n} \frac{(-1)^{m+1}}{m} = \log 2+\frac{1+(-1)^{n+1}}{n}+\mathcal{O}(n^{-2})$ appear – reuns Sep 03 '17 at 20:40

What happens at the boundary of the disk of convergence?

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