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

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 $$z^n/n$$?

## 1 Answer

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

• Can you please explain why or give some reference? – Dr Potato Jun 6 '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 6 '19 at 18:06
• 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 6 '19 at 18:12