Closed form of series $z^n/n$.

Question

Let $z\in\mathbb{C}$. In other question is answered precisely where $\sum\limits_{n=1}^{\infty} \frac{z^n}{n}$ converges.

I have been looking for an expression of $$\sum\limits_{n=1}^{\infty} \frac{z^n}{n}$$ without infinite sum. I mean, a closed expression of this simple hypergeometric sum, similar to, for example, $$\forall |z|<1 , \ \sum\limits_{n=1}^{\infty} z^n = \frac{1}{1-z}.$$ Is there any closed form of the hypergeometric series $z^n/n$?

score 3 · Accepted Answer · answered Jun 06 '19 at 18:02

3

The sum is $-\log(1-z)$, where $\log$ is the main logarithm.

answered Jun 06 '19 at 18:02

José Carlos Santos

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Can you please explain why or give some reference? – Dr Potato Jun 06 '19 at 18:04
Consider the function $z\mapsto-\log(1-z)$. Obviosuly, it maps $0$ into $0$. And if you differentiate it, you get $\frac1{1-z}=1+z+z^2+z^3+\cdots$. – José Carlos Santos Jun 06 '19 at 18:06
2

Integrate both sides of the equality$$\frac1{1-z}=1+z+z^2+z^3+\cdots$$and you get that$$-\log(1-z)=z+\frac{z^2}2+\frac{z^3}3+\frac{z^4}4+\cdots$$ – José Carlos Santos Jun 06 '19 at 18:12

Closed form of series $z^n/n$.

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