Showing $\sum_{k=1}^{\infty}\frac{1}{k}z^{k-1} = - \displaystyle\frac{\ln(1-z)}{z}$

Question

Given the sum $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k}z^{k-1} = 1 + \frac{z}{2} + \frac{z^2}{3} + ...$

How does one show that this sum is equal to $- \displaystyle\frac{\ln(1-z)}{z}$?

Answer 1

Firstly note that your equation is not true for all $z$. Assuming $z$ is real, we need $z$ be in the interval $[-1,1)$. Furthermore $z$ must be non-zero.

To prove your claim for any real non-zero $z$ in the range $(-1,1)$, consider the well-known formula the infinite geometric progression: $$ \frac{1}{1-z} = 1+ z+ z^2 +z^3 + \ldots $$ (Note that the infinite series on the RHS converges for all $z$ in the aforementioned range).

What do you get when you integrate both sides with respect to $z$?

Answer 2

We start with the basic identity $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$ and integrate both sides to get $-\log(1-z) = \int \sum_n z^n dz = \sum_{n=0}^\infty \int z^n dz = \sum_{n=1}^\infty \frac{z^n}{n} = z \sum_{n=1} \frac{z^{n-1}}{n}$ as needed.