The sum originated from this question and I found (experimentally) two closed form cases:
$$\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k)=(-1)^{n+1} H_n$$
$$\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) \frac{1}{2^k}=\frac{1-(-1)^n}{2} \frac{(-1)^{\frac{n-1}{2}} 2^{n-1}}{n} \binom{n-1}{\frac{n-1}{2}}^{-1}$$
Now I'm curious if there's a general closed form for this sum:
$$f_n(x)=\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) x^k$$
Important note: there's a hypergeometric closed sum for this expression, because it originated from a hypergeometric function. So I am searching for elementary closed form (with polynomial, rational and maybe algebraic functions involved).
If we separate it in two parts, we have:
$$f_n (x) =(-1)^n H_n P_n (2x-1)-\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} H_k x^k$$
Which means that the problem reduces to finding:
$$g_n(x)=\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} H_k x^k$$
If we represent the harmonic numbers as integrals, we have:
$$g_n(x)=(-1)^n\int_0^1 \frac{dt}{1-t} \left( P_n(2x-1)-P_n(2 x t-1) \right) $$
Now I guess we need to look up some integrals with Legendre polynomials.
Though this integral explains why $x=1/2$ is a special case.
One of the reasons for my curiosity is this identity for $x>1$:
$$\color{blue}{\log x =-2 \lim_{n \to \infty} \frac{\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) \frac{x^k}{(x-1)^k}}{(-1)^n P_n \left(\frac{x+1}{x-1} \right)}}$$
The limit converges very fast, especially for large $x$.
Compare:
$$F(x,n)=-2 \frac{\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) \frac{x^k}{(x-1)^k}}{(-1)^n P_n \left(\frac{x+1}{x-1} \right)}$$
$$G(x,n)= 2 \sum_{k=0}^n \frac{1}{2k+1} \left(\frac{x-1}{x+1} \right)^{2k+1}$$
The limit performs orders of magnitude better (though it's more difficult to evaluate for the same number of terms clearly).
from Hot Weekly Questions - Mathematics Stack Exchange
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