Binomial coefficient sum over top index

Question

I am trying to evaluate a sum over binomial coefficients which is giving me some problems. Specifically I want to calculate:

$$\sum_{r=0}^{c-1}\binom{r+n}{n}\frac{1}{c-r}$$

My main thought was to convert the fraction here into:

$$\int_0^1 x^{c-r}dx,$$ move the integral out of the sum, alongside the $x^c$ and then attempt to rewrite as some closed form function. I however, cannot see what the generating function should be.

Note, my aim here is to avoid having a sum - some product of binomial coefficients would be ideal but I obviously do not know if this exists!

Any help on summing this would be greatly appreciated.

Answer 1

I could not find a nice closed form; however, I have previously computed the generating function $$ \begin{align} f_m(x) &=\sum_{n=1}^\infty\sum_{k=1}^n\frac{\binom{n-k}{m}}{k}x^n\tag1\\ &=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{\binom{n-k}{m}}{k}x^n\tag2\\ &=\sum_{k=1}^\infty\frac{x^k}k\sum_{n=0}^\infty\binom{n}{m}x^n\tag3\\ &=\sum_{k=1}^\infty\frac{x^k}k\sum_{n=0}^\infty(-1)^{n-m}\binom{-m-1}{n-m}x^n\tag4\\ &=\sum_{k=1}^\infty\frac{x^{\color{#C00}{k}+\color{#090}{m}}}{\color{#C00}{k}}\color{#00F}{\sum_{n=0}^\infty(-1)^{n}\binom{-m-1}{n}x^n}\tag5\\ &=\frac{\color{#090}{x^m}}{\color{#00F}{(1-x)^{m+1}}}\color{#C00}{\log\left(\frac1{1-x}\right)}\tag6 \end{align} $$ Explanation:
$(1)$: definition
$(2)$: switch order of summation
$(3)$: substitute $n\mapsto n+k$
$(4)$: negative binomial coefficients
$(5)$: substitute $n\mapsto n+m$
$(6)$: $\sum\limits_{n=0}^\infty(-1)^{n}\binom{-m-1}{n}x^n=\frac1{(1-x)^{n+1}}$ and $\sum\limits_{k=1}^\infty\frac{x^k}k=\log\left(\frac1{1-x}\right)$

We can apply $(6)$ to get $$ \begin{align} \sum_{r=0}^{c-1}\binom{r+n}n\frac1{c-r} &=\sum_{r=1}^c\binom{c-r+n}n\frac1r\tag7\\ &=\sum_{r=1}^{c+n}\binom{c+n-r}n\frac1r\tag8\\ &=\left[x^{c+n}\right]\frac{x^n}{(1-x)^{n+1}}\log\left(\frac1{1-x}\right)\tag9\\[3pt] &=\left[x^c\right]\frac1{(1-x)^{n+1}}\log\left(\frac1{1-x}\right)\tag{10} \end{align} $$ Explanation:
$\phantom{1}{(7)}$: substitute $r\mapsto c-r$
$\phantom{1}{(8)}$: the terms with $r\in[c+1,c+n]$ are $0$
$\phantom{1}{(9)}$: apply $(6)$
$(10)$: $\left[x^{c+n}\right]x^nf(x)=\left[x^c\right]f(x)$

Thus, $(10)$ gives the generating function for the sums.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{r = 0}^{c - 1}{r + n \choose n}{1 \over c - r}} \\[2mm] = &\ \bracks{z^{c}}\sum_{\ell = 0}^{\infty}z^{\ell} \sum_{r = 0}^{\ell - 1}{r + n \choose n} {1 \over \ell - r} \\[5mm] = &\ \bracks{z^{c}}\sum_{r = 0}^{\infty}{r + n \choose n} \sum_{\ell = r + 1}^{\infty}\,\,{z^{\ell} \over \ell - r} \\[5mm] = &\ \bracks{z^{c}}\sum_{r = 0}^{\infty} {r + n \choose n}z^{r}\ \underbrace{\sum_{\ell = 1}^{\infty}\,\, {z^{\ell} \over \ell}}_{\ds{-\ln\pars{1 - z}}} \\[5mm] = &\ -\bracks{z^{c}}\ln\pars{1 - z}\ \times \\[2mm] &\ \sum_{r = 0}^{\infty} {\bracks{-r - n} + r - 1 \choose r} \pars{-1}^{r}\,z^{r} \\[5mm] = &\ -\bracks{z^{c}}\ln\pars{1 - z} \sum_{r = 0}^{\infty} {- n - 1 \choose r}\pars{-z}^{r} \\[5mm] = &\ -\bracks{z^{c}}\ln\pars{1 - z}\pars{1 - z}^{-n - 1} \\[5mm] = &\ -\bracks{z^{c}}\bracks{\nu^{1}} \pars{1 - z}^{\nu -n - 1} \\[5mm] = &\ -\bracks{\nu^{1}} {\nu - n - 1 \choose c}\pars{-1}^{c} \\[5mm] = &\ -\bracks{\nu^{1}}{-\nu + n + 1 + c - 1 \choose c} \pars{-1}^{c}\pars{-1}^{c} \\[5mm] = &\ -\bracks{\nu^{1}}{c + n - \nu \choose c} \\[5mm] = &\ \left.{n + c - \nu \choose c} \pars{H_{n + c - \nu}\ -\ H_{n - \nu}} \right\vert_{\ \nu\ =\ 0} \\[5mm] = &\ \bbx{{n + c \choose c} \pars{H_{n + c}\ -\ H_{n}}} \\ & \end{align}