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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 $z=(y-1)$

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I don't think that there exists a closed form. I don't know if you would allow the solution obtained from wolfram. It uses hypergeometric functions – Amr Nov 24 '12 at 16:33
@Amr, what is the solution from wolfram? – lab bhattacharjee Nov 24 '12 at 16:35
Even when $y = 1$, where your sum reduces to the famous harmonic number, there is no known simple closed form. – Sangchul Lee Nov 24 '12 at 16:56
up vote 1 down vote accepted

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 $$\begin{eqnarray*} \sum_{1\le r\le n}\frac{y^r}{r} &=& -\log(1-y) - \frac{y^{n+1}}{n+1} \, {}_2 F_1(n+1,1;n+2;y). \end{eqnarray*}$$

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