Show $\sum_{k=0}^\infty \frac{1}{(k+a)\binom{n+k}{k}}=\frac{-\,n!}{(1-a)_n}\Big(\gamma+\psi(a)+\sum_{k=1}^{n-1}\frac{1}{k}\frac{(1-a)_k}{k!}\Big)$

Question

The conjectured identity of the title, $$(C)\quad\sum_{k=0}^\infty \frac{1}{(k+a)\binom{n+k}{k}}=\frac{-\,n!}{(1-a)_n}\Big(\gamma+\psi(a)+\sum_{k=1}^{n-1}\frac{1}{k}\frac{(1-a)_k}{k!}\Big)$$ arose in an attempt to generalize a specific case of this identity for $a=1/2.$ (The symbol $(a)_k$ is the Pochhammer symbol.) The specific identity I proved was $$\sum_{k=0}^\infty \frac{1}{(k+1/2)\binom{n+k}{k}}=\frac{n!}{(1/2)_n} \sum_{k=n}^\infty \frac{1}{k} \frac{(1/2)_k}{k!}=\frac{2^{2n}}{\binom{2n}{n}}\big(2\log{2} - \sum_{k=1}^{n-1}\frac{2^{-2k}}{k}\binom{2k}{k}\big)$$ This form is suggestive of $(C),$ for we can sum from 1 to $\infty$ and subtract the series tail and use the known summation, $$\sum_{k=1}^\infty \frac{1}{k} \frac{(1-a)_k}{k!}=-\gamma - \psi(a),$$ the two terms being Euler's constant and the digamma function, respectively. The proof I prefer is one that doesn't require a knowledge of the RHS of $(C).$ For example, I'd prefer not to have it shown that both sides obey the same recursion relationship with a boundary condition.

Answer 1

The sum is an hypergeometric series : $$\sum_{k=0}^\infty \frac{1}{(k+a)\binom{n+k}{k}}=\sum_{k=0}^\infty \frac{n!k!}{(k+a)(k+n)!}=n!\sum_{k=0}^\infty \frac{\Gamma(k+1)}{(k+a)\Gamma(n+k+1)} $$ From the properties of the Hypergeometric function : http://mathworld.wolfram.com/HypergeometricFunction.html $$_3F_2(1,1,a;a+1,n+1;x)=\sum_{k=0}^\infty \frac{\Gamma(k+1)\Gamma(k+1)\Gamma(k+a)\Gamma(a+1)\Gamma(n+1)}{\Gamma(1)\Gamma(1)\Gamma(a)\Gamma(k+a+1)\Gamma(k+n+1)}\frac{x^k}{k!} \\=a\:n!\sum_{k=0}^\infty \frac{\Gamma(k+1)}{(k+a)\Gamma(k+n+1)}x^k$$ With $x=1$ :

$$\sum_{k=0}^\infty \frac{\Gamma(k+1)}{(k+a)\Gamma(k+n+1)}=\frac{1}{an!}\,_3F_2(1,1,a;a+1,n+1;1)$$

$$\sum_{k=0}^\infty \frac{1}{(k+a)\binom{n+k}{k}}=\frac{1}{a}\,_3F_2(1,1,a;a+1,n+1;1)$$ A lot of relationships with other series and/or functions can be found in : http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/

Also $\sum_{k=1}^{n-1} \frac{(1-a)_k}{k!k} = \frac{1}{\Gamma(1-a)}\sum_{k=1}^{n-1} \frac{\Gamma(k+1-a)}{k!k} $ can be expressed in term of hypergeometric function.

With the help of WolframAlpha :

Replacing $x$ by $(1-a)$ and changing $n$ to $n-1$ leads to : $$\sum_{k=1}^{n-1} \frac{(1-a)_k}{k!k}= -\frac{\Gamma(n+1-a)}{\Gamma(1-a)n^2(n-1)!}\,_3F_2(1,n,n+1-a;n+1,n+1;1) - \frac{\Gamma(2-a)}{\Gamma(1-a)(1-a)}(\psi(a)+\gamma).$$

$$\gamma+\psi(a)+\sum_{k=1}^{n-1} \frac{(1-a)_k}{k!k}= -\frac{\Gamma(n+1-a)}{\Gamma(1-a)n^2(n-1)!}\,_3F_2(1,n,n+1-a;n+1,n+1;1).$$

$$\frac{-n!}{(1-a)_n}\left(\gamma+\psi(a)+\sum_{k=1}^{n-1} \frac{(1-a)_k}{k!k} \right) = \frac{1}{n}\,_3F_2(1,n,n+1-a;n+1,n+1;1) $$ And with the relationship between the two hypergeometric functions at argument=$1$ : $$\,_3F_2(1,n,n+1-a;n+1,n+1;1) = \frac{n}{a}\,_3F_2(1,1,a;a+1,n+1;1) \tag 1$$ $$\frac{-n!}{(1-a)_n}\left(\gamma+\psi(a)+\sum_{k=1}^{n-1} \frac{(1-a)_k}{k!k} \right) =\frac{1}{a}\,_3F_2(1,1,a;a+1,n+1;1) $$ With the relationship found above $\sum_{k=0}^\infty \frac{1}{(k+a)\binom{n+k}{k}}=\frac{1}{a}\,_3F_2(1,1,a;a+1,n+1;1)$ one obtains :

$$\sum_{k=0}^\infty \frac{1}{(k+a)\binom{n+k}{k}}=\frac{-n!}{(1-a)_n}\left(\gamma+\psi(a)+\sum_{k=1}^{n-1} \frac{(1-a)_k}{k!k} \right)$$ which is the expected result.

Note : The most likely the relation (1) can be found in the literature. But I found no reference with only a too short search. On the other hand, my attempt to prove it analytically is not sufficiently rigorous to publish it.