Two problems of binomial sums involving harmonic numbers

Two open problems recently proposed by Xi and Luo (Adv. Diﬀer. 2021) are resolved by evaluating explicitly three binomial sums involving harmonic numbers, that are realized mainly by utilizing the generating function method and symmetric functions. 05A10; 05A19; 11B65

The harmonic numbers of higher order are given by H (λ) 0 = 1 and H (λ) n = n k=1 1 k λ for n, λ ∈ N.
In order to reduce lengthy expressions, we shall employ the notations of elementary and complete symmetric functions. For a finite set S of real numbers, we define these functions by 0 (x|S) = 0 (x|S) ≡ 1 and We shall also need the signless Stirling numbers of the first kind (see [6]) which are determined by the connection coefficient of expanding the shifted factorials into monomials There exist numerous summation formulae involving harmonic numbers (cf. [1-3, 7, 8]).
In a recent paper [9], Xi The first binomial sum in Problem I can easily be evaluated by the Chu-Vandermonde convolution formula as follows: As the primary motivation, the aim of the present paper is to resolve these problems and evaluate the remaining three sums explicitly in the following theorems.
Theorem 1 Let x be an indeterminate. Then for m, n ∈ N 0 , the following algebraic identity holds: We remark that when m > n, this theorem evaluates the second sum in Problem I by determining the polynomial part of the rational function explicitly as in the last line, which vanishes for m ≤ n, instead.
Theorem 2 Let x be an indeterminate. Then for m, n, λ ∈ N 0 , the following algebraic identity holds: Theorem 3 Let x be an indeterminate. Then for m, n, λ, ρ ∈ N 0 , the following algebraic identity holds: The rest paper will be organized as follows. In the next section, we shall prove Theorem 1 by determining explicitly the polynomial part of a rational function when its numerator degree is greater than that of the denominator. Then Theorems 2 and 3 will be shown in Sect. 3 by establishing two analytical formulae of the derivatives of higher order for a polynomial function of the rising factorial and its reciprocal. The informed reader will notice that by employing symmetric functions and , several involved expressions become simpler than those appearing in [9], where the Bell polynomials were employed.

Proof of Theorem 1
Observe that the rational function below can be decomposed into partial fractions where P m n (x) is a polynomial of degree mn -1 in x which reduces to zero when m ≤ n, and the coefficients A k are determined by the limits Therefore, we have found the equality By scaling down m and then making use of we can rewrite the last equality as Evaluating the last sum by means of the Chu-Vandemonde formula and then comparing the resultant expression with (4), we get the following recurrence relation: In order to find an explicit expression for P m n (x), let Q m := P m+n n (x). Then the equality corresponding to (5) becomes It is routine to figure out the initial values Q 0 = 0 and Q 1 = (-1) n+1 n+1 . Then we can manipulate the generating function (-1) n m + n -1 n y m+n m + n .
By differentiating the last equation with respect to y, (-1) n m + n -1 n y m+n-1 , and then evaluating the binomial series on the right, we find, after some simplification, that Q(y) satisfies the following differential equation: It is trivial to check that the corresponding homogeneous equation where is an arbitrary constant. When := (y) is considered as a function of y, substituting the above solution into (7) gives rise to (y) = (-1) n+1 y n (1y) -x-n-1 .
Therefore, we have the integral representation Define for simplicity According to integration by parts, we can calculate J n as follows: By means of the induction principle, we can show that which is equivalent to the expression (-1) n+k n k x + n k+1 y n-k (1y) k-x-n .
Substituting this into (8), we obtain the explicit generating function Extracting the coefficient of y m+n across the last equation yields By reformulating the last sum with respect to k as n k=0 m + n nk we find finally the binomial expression This gives consequently the desired formula stated in Theorem 1: (-1) n+k m n + k (x + n + 1) k-1 (n + 1) k .

Proofs of Theorems 2 and 3
For the derivative operator D with respect to x, we have the following analytical formulae of higher order derivatives: . (10) The first one in (9) can be evaluated easily by induction on n. In order to prove the second one in (10), define where the function G n remains to be determined with the initial values G 0 = 1 and G 1 = -1 (x|S).
Then by making use of the Leibniz rule, we have which leads us to the binomial recursion In order to find an explicit expression for G λ , we examine the exponential generating function defined by According to (12), its derivative with respect to y can be expressed as We therefore get the differential equation whose solution is given by the exponential function Evaluating the last sum with respect to k explicitly as we find the simplified generating function By extracting the coefficient of y n , we confirm the formula (10) as follows: G n = n! y n G(y) = n!(-1) n n (x|S).
Substituting the above three expressions into the equality of Theorem 1 and then making some simplifications, we find the algebraic identity in Theorem 2.