Like Arithmetico-geometric series, is there anyway to calculate in closed form of Geomtrico-harmonic series like $\sum_{1\le r\le n}\frac{y^r}r$ where $n$ is finite.
We know if $n\to \infty,$ the series converges to $-\log(1+y)$ for $-1\le y<1$
The way I have tried to address it is as follows:
we know, $\sum_{0\le s\le n-1}y^s=\frac{y^n-1}{y-1}$
Integrating either sides wrt $y$, we get $\sum_{1\le r\le n}\frac{y^r}r=\int \left(\frac{y^n-1}{y-1}\right) dy$
but how to calculate this integral in the closed form i.e., without replacement like $z=(y-1)$