Geometrico-Harmonic Progression 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 $z=(y-1)$
 A: It doesn't appear that a simple closed form for this sum exists.
As @sos440 suggests in the comments, we shouldn't expect to find one.
As noted by @Amr, the sum is related to the hypergeometric function.
We have 
$$\begin{eqnarray*}
\sum_{1\le r\le n}\frac{y^r}{r}
    &=& \sum_{r=1}^\infty\frac{y^r}{r} 
    - \sum_{r=n+1}^\infty\frac{y^r}{r} \\
    &=& -\log(1-y) - y^{n+1}\sum_{k=0}^\infty \frac{y^k}{n+k+1}.
\end{eqnarray*}$$
The ratio of successive terms in the hypergeometric series
$${}_2 F_1(a,b;c;y) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} 
    \sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)}{\Gamma(c+k)k!} y^k$$
is 
$$\frac{(a+k)(b+k)y}{(c+k)(k+1)}.$$
Thus, the sum above is a hypergeometric series with $a = n+1$, $b=1$, and $c = n+2$. 
There is an overall factor of $1/(n+1)$, so 
$$\begin{eqnarray*}
\sum_{1\le r\le n}\frac{y^r}{r}
    &=& -\log(1-y) 
    - \frac{y^{n+1}}{n+1} \, {}_2 F_1(n+1,1;n+2;y).
\end{eqnarray*}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{1\ \leq\ r\ \leq\ n}{y^{r} \over r} & =
\overbrace{\sum_{r = 1}^{\infty}{y^{r} \over r}}
^{\ds{-\ln\pars{1 - y}}}\ -\
\sum_{r = n + 1}^{\infty}{y^{r} \over r} =
-\ln\pars{1 - y} -
y^{n + 1}\sum_{r = 0}^{\infty}{y^{r} \over r + n + 1}
\\[5mm] & =
\bbx{-\ln\pars{1 - y} - y^{n + 1}\,\Phi\pars{y,1,n + 1}}
\end{align}

$\ds{\Phi}$ is the
  Lerch Trascendent Function or LerchPhi Function.

