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I already found 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 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 $z=1$ the series converges.

Now how can I proceed? Please help me.

Thank you in advance.

Community
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  • The series $$\sum_{n=1}^{\infty} \frac {(-1)^n} {n}.$$ Which is convergent by Leibnitz's test. –  Sep 03 '17 at 20:29
  • For the real arguments $\pm 1$, yes. But this is a complex domain problem. – Oscar Lanzi Sep 03 '17 at 20:55
  • A ring of doom? I think there might be infinitely may singularities on the boundary. – Oscar Lanzi Sep 03 '17 at 20:58
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    Exact duplicate of question asked by other user 6 hours ago https://math.stackexchange.com/questions/2415413/what-happens-at-the-boundary-of-the-disk-of-convergence#comment4988856_2415413 – reuns Sep 03 '17 at 21:16

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"I found that the radius of convergence to be 1 and hence the disk of convergence is $|z|=1$" That's not a disk, it is a circle. I presume you mean that convergence is clear for $|z|< 1$. What happens on the circle $|z|= 1$ depends on the precise value of $z$.

"Now I found at $z=1$ the series converges." Okay, the series in question is $\sum_{n=1}^\infty \frac{(-1)^n}{n}z^{n(n+1)}$. At z= 1 that is the alternating series $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ which converges because $\frac{1}{n}$ is descending.

Any point on the circle $|z|= 1$ can be written $z= e^{i\theta}$ for $0\le \theta< 2\pi$. So the series is $\sum_{n=1}^\infty \frac{(-1)^n}{n} e^{n(n+1)i\theta}$

Boaz Moerman
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