TITLE: Is there a close form for $g(a,b,n)=\sum\limits_{k=0}^{n}\binom{n}{k}\frac{1}{ak+b}$?
QUESTION [5 upvotes]: 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 $g(a,b,n)$?

REPLY [2 votes]: 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.