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I've been given the following series complex series: $$\sum_{n=1}^\infty \frac{(-1)^n}{n}z^n $$ for $z\in\mathbb{C}$. It is easy to see( by the ratio test) that the series converges when $|z|< 1$ and diverges when $|z|>1$. I need to see where at the boundary this series converges and where it diverges.

The nth term goes to $0$ when $|z|=1$, which doesn't give information to us. If $|z|=1$, then there exists a $\theta\in[0,2\pi]$ such that $z= e^{i\theta}$. So, the series has real part: $$\sum_{n=1}^\infty \frac{(-1)^n}{n}\cos(n\theta) $$ and imaginary part: $$\sum_{n=1}^\infty \frac{(-1)^n}{n}\sin(n\theta) $$ But these are alternating series to which the Leibnitz criterion cannot be applied ( the generating sequence isn't decreasing, nor positive). I don't really know how to proceed from here. Can you help me? Also tell if you know an easier way of checking this.

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    $\begingroup$ Are you comfortable dealing with the series that has no $(-1)^n$? If so notice that $(-1)^n=e^{i\pi n}$ so it can be asborbed in the exponential by the change of variable $\theta$ goes into $\theta + \pi$ where $z=re^{i\theta}$ $\endgroup$
    – Conrad
    Oct 12, 2019 at 14:28

2 Answers 2

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If $z=-1$, then the series diverges, since it is just the harmonic series. Otherwise, it converges, by Dirichlet's test. There is no need to conider the real part and the imaginary part separately.

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  • $\begingroup$ Can you specify it a little bit? I don't see how to apply that test. Which part has partial sums bounded and which part is decreasing to 0 ? $\endgroup$
    – Seven
    Oct 12, 2019 at 14:17
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    $\begingroup$ $$\lvert-z+z^2-z^3+z^4-\cdots\pm z^n\rvert=\left\lvert\frac{-z\mp z^{n+1}}{1+z}\right\rvert\leqslant\frac2{\lvert1+z\rvert}.$$ $\endgroup$
    – José Carlos Santos
    Oct 12, 2019 at 14:22
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This series can be summed in a closed form. We have $$\frac{1}{1-z}=\sum_{n\ge0}z^n.$$ Then $$\int\frac{1}{1-z}dz=-\log(1-z)=\sum_{n\ge0}\frac{z^{n+1}}{n+1}=\sum_{n\ge1}\frac{z^{n}}{n}.$$ So your series is just $-\log(1+z)$.

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