Binomial coefficient sum over top index 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.
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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)}
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\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}
