Does $\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + cs}$ have a closed-form expression?

Question

I'm wondering whether $$ \sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + c \cdot s}, $$ with $\delta \in (0,1)$ and $c\ge 0$ has a closed-form expression.

While I don't know of a general closed-form I can think if a special case:

Setting $c=0$ we obtain $$ \sum^\infty_{s=1}\delta^{s-1}=\frac{1}{1-\delta}, $$ which is the geometric series. As for the more general case $c\neq 0$ I'm not sure.

Answer 1

For $|x^c|< 1$, consider the geometric series $\sum\limits_{n=0}^{\infty}x^{nc}$. Then we know $$\frac{1}{1-x^c}=\sum\limits_{n=0}^{\infty}x^{nc}.$$ So \begin{align*} \frac{x^c}{1-x^c}& =\sum_{n=0}^{\infty}x^{nc+c}\\ \int\frac{x^c}{1-x^c} \, dx& =\sum_{n=0}^{\infty}\frac{x^{nc+c+1}}{1+c(n+1)}\\ \int\frac{x^c}{1-x^c} \, dx& =x^{c+1}\sum_{n=0}^{\infty}\frac{x^{nc}}{1+c(n+1)}. \end{align*} Now let $\delta=x^c$ (with $c \neq 0$), so we have \begin{align*} \frac{1}{c}\int\frac{\delta^{1/c}}{1-\delta} \, d\delta& =\delta^{\frac{c+1}{c}}\sum_{n=0}^{\infty}\frac{\delta^{n}}{1+c(n+1)}\\ \color{red}{\frac{1}{c\delta^{\frac{c+1}{c}}}\int\frac{\delta^{1/c}}{1-\delta} \, d\delta}& =\sum_{n=0}^{\infty}\frac{\delta^{n}}{1+c(n+1)}. \end{align*} So if we can solve the integral then we can get some closed form. However solving this integral may not be that straightforward.

Answer 2

Picking up where @Anurag left off we have $$ \begin{align*} \sum_{n=1}^{\infty}\frac{\delta^{n-1}}{1+cn} &=\frac{1}{c\delta^{\frac{c+1}{c}}}\int\frac{\delta^{1/c}}{1-\delta} \,\mathrm d\delta\\ &=\frac{1}{c\delta^{\frac{c+1}{c}}}\int_0^\delta\frac{t^{1/c}}{1-t} \,\mathrm dt\\ &\overset{t=\delta z}{=}\frac{1}{c}\int_0^1\frac{z^{1/c}}{1-\delta z} \,\mathrm dz\\ &=\frac{1}{c}\int_0^1\frac{z^{(1+1/c)-1}(1-z)^{(2+1/c)-(1+1/c)-1}}{1-\delta z} \,\mathrm dz\\ &=\frac{1}{c(1+1/c)}(1+1/c)\int_0^1\frac{z^{(1+1/c)-1}(1-z)^{(2+1/c)-(1+1/c)-1}}{1-\delta z} \,\mathrm dz\\ &=\frac{1}{c+1}\frac{\Gamma(2+1/c)}{\Gamma(1+1/c)\Gamma(1)}\int_0^1\frac{z^{(1+1/c)-1}(1-z)^{(2+1/c)-(1+1/c)-1}}{1-\delta z} \,\mathrm dz, \end{align*} $$ which according to DLMF 15.6.1 equals $$ \sum_{n=1}^{\infty}\frac{\delta^{n-1}}{1+cn}=\frac{1}{c+1}{_2F_1}\left({1,1+\frac{1}{c}\atop 2+\frac{1}{c}};\delta\right), $$ with ${_2F_1}(\cdot)$ being the (Gauss) hypergeometric function.

Note that the series representation converges for $|\delta|<1$ and $c\in\Bbb C\setminus\{-1,-2,\dots\}$, whereas the hypergeometric representation can be extended to $\delta\in\Bbb C$ via analytic continuation and thus is a more general form.

Furthermore, using this relationship we can also express this result in terms of the Lerch Transcendent as $$ \sum_{n=1}^{\infty}\frac{\delta^{n-1}}{1+cn}=\frac{1}{c}\Phi(\delta,1,1+\tfrac{1}{c}). $$