Is there a close form for $g(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{ak+b}$?

Question

We can be sure, that for $a>0$, $b>0$ $$f(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{ak+b}=\frac{(an)!^{(a)}}{(an+b)!^{(a)}}$$ where $(an+b)!^{(a)}$ denotes multifactorial: $(n)!^{(1)}=n!$, $(2n)!^{(2)}=(2n)!!$, etc.

But if we make a little change $$g(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{ak+b}$$ we have $$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}=\frac{2^{n+1}-1}{n+1}$$ $$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+2}=\frac{n2^{n+1}+1}{(n+1)(n+2)}$$ in general $$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+2c}=\frac{n(2c-1)!}{n^{\overline {2c+1}}}\left(1+2^{n+1}\sum\limits_{m=1}^{c}\frac{n^{\overline {2m-1}}}{(2m-1)!}\right)$$ $$\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{k+2c-1}=\frac{n(2c-2)!}{n^{\overline {2c}}}\left(-1+2^{n+1}\sum\limits_{m=1}^{c}\frac{n^{\overline {2m-2}}}{(2m-2)!}\right)$$ Is there a close form for $g(a,b,n)$?

Answer 1

We start by trying to prove the first closed form given to see if a pattern does emerge. We use with $c$ a positive integer

$${n+c\choose n} \sum_{k=0}^n {n\choose k} \frac{1}{k+c}$$

Now

$${n+c\choose n} {n\choose k} = \frac{(n+c)!}{ (c)! \times k! \times (n-k)!} = {n+c\choose k+c} {k+c\choose k}.$$

Hence we have for the sum

$$\sum_{k=0}^n {n+c\choose k+c} {k+c\choose k} \frac{1}{k+c} = \frac{1}{c} \sum_{k=0}^n {n+c\choose k+c} {k+c-1\choose c-1}.$$

This is

$$\frac{1}{c} \sum_{k=0}^n {k+c-1\choose c-1} [z^{n-k}] \frac{1}{(1-z)^{k+c+1}} = \frac{1}{c} \sum_{k=0}^n {k+c-1\choose c-1} [z^{n}] z^k \frac{1}{(1-z)^{k+c+1}}.$$

Here we get no contribution to $[z^n]$ when $k\gt n$ so we may continue with

$$\frac{1}{c} [z^n] \frac{1}{(1-z)^{c+1}} \sum_{k\ge 0} {k+c-1\choose c-1} z^k \frac{1}{(1-z)^{k}} \\ = \frac{1}{c} [z^n] \frac{1}{(1-z)^{c+1}} \frac{1}{(1-z/(1-z))^c} \\ = \frac{1}{c} [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^c}.$$

This is

$$\frac{1}{c} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{1-z} \frac{1}{(1-2z)^c} \\ = \frac{(-1)^{c+1}}{c 2^c} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{z-1} \frac{1}{(z-1/2)^c}.$$

With residues summing to zero we can evaluate this using the residues at $z=1$, $z=1/2$ and $z=\infty.$ We get for $z=1$ the residue

$$\frac{(-1)^{c+1}}{c}.$$

For the residue at infinity we find

$$- \frac{(-1)^{c+1}}{c 2^c} \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1}{(1/z)^{n+1}} \frac{1}{1/z-1} \frac{1}{(1/z-1/2)^c} \\ = - \frac{(-1)^{c+1}}{c 2^c} \mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1} \frac{z}{1-z} \frac{z^c}{(1-z/2)^c} \\ = - \frac{(-1)^{c+1}}{c 2^c} \mathrm{Res}_{z=0} z^{n+c} \frac{1}{1-z} \frac{1}{(1-z/2)^c} = 0.$$

This also follows by inspection. The residue at $z=1/2$ requires the use of Leibniz' rule as in

$$\frac{1}{p!} \left( \frac{1}{z^{n+1}} \frac{1}{z-1} \right)^{(p)} = \frac{1}{p!} \sum_{q=0}^p {p\choose q} \frac{(-1)^q (n+q)!}{n! z^{n+1+q}} (-1)^{p-q} \frac{(p-q)!}{(z-1)^{p-q+1}} \\ = (-1)^p \sum_{q=0}^p {n+q\choose q} \frac{1}{z^{n+1+q}} \frac{1}{(z-1)^{p-q+1}}.$$

We set $p=c-1$ and $z=1/2$ and restore the factor in front to get for the residue

$$\frac{(-1)^{c+1}}{c 2^c} (-1)^{c-1} \sum_{q=0}^{c-1} {n+q\choose q} \frac{1}{(1/2)^{n+1+q}} \frac{(-1)^{c-q}}{(1/2)^{c-q}} \\ = \frac{(-1)^c 2^{n+1}}{c} \sum_{q=0}^{c-1} {n+q\choose q} (-1)^q.$$

Collecting everything we thus obtain

$$\bbox[5px,border:2px solid #00A000]{ \sum_{k=0}^n {n\choose k} \frac{1}{k+c} = {n+c\choose c}^{-1} \frac{(-1)^c}{c} \left(1-2^{n+1} \sum_{q=0}^{c-1} {n+q\choose q} (-1)^q\right).}$$

This is an improvement in the sense that if $n$ is the variable and $c$ is the constant then we have replaced the sum in $n$ terms (variable) by a sum in $c$ terms (fixed) of polynomials in $n.$ We can make this more explicit by writing

$$\sum_{q=0}^{c-1} {n+q\choose q} (-1)^q = \sum_{q=0}^{c-1} \frac{(-1)^q}{q!} \sum_{p=0}^q n^p {q+1\brack p+1} \\ = \sum_{p=0}^{c-1} n^p \sum_{q=p}^{c-1} \frac{(-1)^q}{q!} {q+1\brack p+1}.$$

We find

$$\bbox[5px,border:2px solid #00A000]{ \sum_{k=0}^n {n\choose k} \frac{1}{k+c} = {n+c\choose c}^{-1} \frac{(-1)^c}{c} \left(1-2^{n+1} \sum_{p=0}^{c-1} n^p \sum_{q=p}^{c-1} \frac{(-1)^q}{q!} {q+1\brack p+1}\right).}$$

With this last result we obtain closed forms for fixed $c$, e.g. for $c=5$ it yields

$$\frac{-24+2^{n+1} (n^4+6n^3+23n^2+18n+24)} {(n+5)\times\cdots\times (n+1)}.$$

Addendum. With the purpose of matching conjectures by OP we write

$$\sum_{q=0}^{c-1} {n+q\choose q} (-1)^q = \sum_{q=0}^{c-1} {n+q\choose q} (-1)^q [z^{c-1}] \frac{z^q}{1-z} \\ = [z^{c-1}] \frac{1}{1-z} \sum_{q\ge 0} {n+q\choose q} (-1)^q z^q = [z^{c-1}] \frac{1}{1-z} \frac{1}{(1+z)^{n+1}} \\ = (-1)^{c-1} [z^{c-1}] \frac{1}{1+z} \frac{1}{(1-z)^{n+1}} = (-1)^{c-1} [z^{c-1}] \frac{1}{1-z^2} \frac{1}{(1-z)^{n}}.$$

With $c=2d+1$ where $d\ge 0$ this becomes

$$[z^{2d}] \frac{1}{1-z^2} \frac{1}{(1-z)^{n}} = \sum_{q=0}^d {2q+n-1\choose 2q}$$

and when $c=2d$ where $d\ge 1$ it becomes

$$- [z^{2d-1}] \frac{1}{1-z^2} \frac{1}{(1-z)^{n}} = - \sum_{q=0}^{d-1} {2q+n\choose 2q+1}.$$

We thus obtain in the first case the closed form

$$\bbox[5px,border:2px solid #00A000]{ {n+2d+1\choose 2d+1}^{-1} \frac{1}{2d+1} \left(-1+2^{n+1} \sum_{q=0}^d {2q+n-1\choose 2q} \right)}$$

and in the second case

$$\bbox[5px,border:2px solid #00A000]{ {n+2d\choose 2d}^{-1} \frac{1}{2d} \left(1+2^{n+1} \sum_{q=0}^{d-1} {2q+n\choose 2q+1} \right).}$$

These two confirm the conjectures by OP.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\Large\mrm{f}\pars{a,b,n} = {\large ?}.}$

\begin{align} \mrm{f}\pars{a,b,n} & \equiv \sum_{k = 0}^{n}{n \choose k}{\pars{-1}^{k} \over ak + b} = {1 \over a}\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k} \int_{0}^{1}t^{k + b/a - 1}\,\dd t \\[5mm] & = {1 \over a}\int_{0}^{1}t^{b/a - 1} \sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k}\,\dd t = {1 \over a}\int_{0}^{1}t^{b/a - 1}\pars{1 - t}^{n}\,\dd t \\[5mm] & = \bbx{{1 \over a}\,{\Gamma\pars{b/a}\Gamma\pars{n + 1} \over \Gamma\pars{b/a + n + 1}}\,,\qquad\Re\pars{b \over a} > 0} \end{align}

---

$\ds{\Large\mrm{g}\pars{a,b,n} = {\large ?}.}$

In following the $\textsf{'above procedure'}$, I'll arrive to \begin{align} \mrm{g}\pars{a,b,n} & \equiv \sum_{k = 0}^{n}{n \choose k}{1 \over ak + b} = {1 \over a}\int_{0}^{1}t^{b/a - 1}\pars{1 + t}^{n}\,\dd t \,\,\,\stackrel{t\ \mapsto\ 1- t}{=}\,\,\, {1 \over a}\int_{0}^{1}\pars{1 - t}^{b/a - 1}\,\pars{2 - t}^{n}\,\dd t \\[5mm] & = {2^{n} \over a}\int_{0}^{1}t^{1 - 1}\pars{1 - t}^{b/a - 1} \,\pars{1 - {1 \over 2}\,t}^{n}\,\dd t \end{align}

which is related to the Euler Type Expression for the Hypergeometric Function $\ds{\mbox{}_{2}\mrm{F}_{1}}$.

Namely, \begin{align} \mrm{g}\pars{a,b,n} & \equiv \sum_{k = 0}^{n}{n \choose k}{1 \over ak + b} = {2^{n} \over a}\,\ \overbrace{\mrm{B}\pars{1,{b \over a}}}^{\ds{a \over b}}\ \mbox{}_{2}\mrm{F}_{1}\pars{-n,1;{b \over a} + 1;{1 \over 2}} \\[5mm] & = \bbx{{2^{n} \over b}\, \mbox{}_{2}\mrm{F}_{1}\pars{-n,1;{b \over a} + 1;{1 \over 2}}\,,\qquad \Re\pars{b \over a} > 0} \end{align}

$\ds{\mrm{B}}$ is the Beta Function.