Vtyjkyuk

The conjectured identity of the title,
$$(C)quadsum_k=0^infty frac1(k+a)binomn+kk=frac-,n!(1-a)_nBig(gamma+psi(a)+sum_k=1^n-1frac1kfrac(1-a)_kk!Big)$$
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
$$sum_k=0^infty frac1(k+1/2)binomn+kk=fracn!(1/2)_n sum_k=n^infty frac1k frac(1/2)_kk!=frac2^2nbinom2nnbig(2log2 - sum_k=1^n-1frac2^-2kkbinom2kkbig)$$
This form is suggestive of $(C),$ for we can sum from 1 to $infty$ and subtract the series tail and use the known summation,
$$sum_k=1^infty frac1k frac(1-a)_kk!=-gamma - psi(a),$$
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.

The sum is an hypergeometric series :
$$sum_k=0^infty frac1(k+a)binomn+kk=sum_k=0^infty fracn!k!(k+a)(k+n)!=n!sum_k=0^infty fracGamma(k+1)(k+a)Gamma(n+k+1) $$
From the properties of the Hypergeometric function : http://mathworld.wolfram.com/HypergeometricFunction.html
$$_3F_2(1,1,a;a+1,n+1;x)=sum_k=0^infty fracGamma(k+1)Gamma(k+1)Gamma(k+a)Gamma(a+1)Gamma(n+1)Gamma(1)Gamma(1)Gamma(a)Gamma(k+a+1)Gamma(k+n+1)fracx^kk! \=a:n!sum_k=0^infty fracGamma(k+1)(k+a)Gamma(k+n+1)x^k$$
With $x=1$ :

$$sum_k=0^infty fracGamma(k+1)(k+a)Gamma(k+n+1)=frac1an!,_3F_2(1,1,a;a+1,n+1;1)$$

$$sum_k=0^infty frac1(k+a)binomn+kk=frac1a,_3F_2(1,1,a;a+1,n+1;1)$$
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)_kk!k = frac1Gamma(1-a)sum_k=1^n-1 fracGamma(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 :
$$sum_k=1^n-1 frac(1-a)_kk!k= -fracGamma(n+1-a)Gamma(1-a)n^2(n-1)!,_3F_2(1,n,n+1-a;n+1,n+1;1) - fracGamma(2-a)Gamma(1-a)(1-a)(psi(a)+gamma).$$

$$gamma+psi(a)+sum_k=1^n-1 frac(1-a)_kk!k= -fracGamma(n+1-a)Gamma(1-a)n^2(n-1)!,_3F_2(1,n,n+1-a;n+1,n+1;1).$$

$$frac-n!(1-a)_nleft(gamma+psi(a)+sum_k=1^n-1 frac(1-a)_kk!k right) = frac1n,_3F_2(1,n,n+1-a;n+1,n+1;1) $$
And with the relationship between the two hypergeometric functions at argument=$1$ :
$$,_3F_2(1,n,n+1-a;n+1,n+1;1) = fracna,_3F_2(1,1,a;a+1,n+1;1) tag 1$$
$$frac-n!(1-a)_nleft(gamma+psi(a)+sum_k=1^n-1 frac(1-a)_kk!k right) =frac1a,_3F_2(1,1,a;a+1,n+1;1) $$
With the relationship found above $sum_k=0^infty frac1(k+a)binomn+kk=frac1a,_3F_2(1,1,a;a+1,n+1;1)$ one obtains :

$$sum_k=0^infty frac1(k+a)binomn+kk=frac-n!(1-a)_nleft(gamma+psi(a)+sum_k=1^n-1 frac(1-a)_kk!k right)$$
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.