A note on some classes of series for the logarithmic function

A diﬀerent and technical proof for a series representation of log s is provided. The mentioned series representation was used to reproduce some Bailey-Borwein-Plouﬀe (BBP) type formulas for a variety of numbers, including some classical BBP type formulas for π . A series representation, similar to that of log s , for log 2 ( s ) is also derived


Introduction
In [2], the authors defined the iterated primitives of the function f (s) = 1  s by I 0 (s) = 1 s and for n ≥ 1.They called I n (s) the Laplacian-Hadamard regularization or the Laplace-Hadamard transform of 1 t n .They also proved that where and , which is the n-th harmonic number.By applying (1) and the Taylor series expansion of the function log s at s = 1, they derived for |s − 1| < 1 or |s − 1| = 1 and n ≥ 2 the following identity: By combining ( 2) and ( 4), the authors in [2] derived the following series representation for log s: where B n (s) is as given in (3).By substituting particular values of s in (5) and considering real and imaginary parts separately, they obtained Bailey-Borwein-Plouffe (BBP) type formulas for a variety of numbers, including the following classical BBP type series for π: (the BBP formula, see [1]), and (see [9]).
For an elementary approach to BBP type formulas for π, the reader is referred to the recent paper [11], and for some related results, the reader is referred to [12].This article provides a completely different and novel proof for (5).A new series representation of type (5) for log 2 (s) is also offered.The proofs are primarily based on the binomial theorem.

Preliminaries
For our purpose, we recall the following definitions and basic properties of some special functions and special numbers.The gamma function is defined by Γ(z) = ∞ 0 u z−1 e −u du (z > 0).The psi function or the digamma function, denoted by ψ(x), is given by ψ(z) = Γ (z) Γ(z) .The polygamma function ψ (n) is defined by The harmonic numbers of order are known in the literature to be When = 1, they reduce to the classical harmonic numbers Throughout this article, Z (Z ) denotes the set of integers less than or equal to (greater than or equal to) some belonging to Z.The following relation between the harmonic numbers and the digamma function is valid: for n ∈ Z ≥0 , where γ = lim (see [13,Section 1.2]).A generalized binomial coefficient s t is given, in terms of the gamma function, by Taking into account that 1 Γ(k) = 0 for k ∈ Z ≤0 , it is clear from this definition that for n, k ∈ Z ≥0 .We now present a series of lemmas, which are used in the next section to prove the main results.Throughout this article, we employ the usual convention that an empty sum is taken to be zero.
Proof.By making the substitution n − k − 1 = k and utilizing (9), one has which proves (11).By employing (10), the identity ( 12) can be proved in the same way as used in the proof of (11).
Lemma 2.3.The following identities hold: Proof.All the identities presented in this lemma are well known in the field (for example, see [7]).
Proof.By the general binomial theorem, for t ∈ (−1, 1) and x ∈ C\Z ≤0 , we have We differentiate both sides of this equation with respect to the variable x.This gives In order to justify termwise differentiation of the left-hand side of (18), it suffices to show that the left-hand side of (19) converges uniformly for x > 1 (see, e.g., [14, p. 231, Theorem 7.14]; see also [10, p. 53, Theorem 2.25]).In terms of convergence, it is enough to show that the case r = 1: converges uniformly on x ∈ (1, M ] for any fixed M > 1.By using (13), we find By using ( 14) and ( 16), we find which leads to By combining ( 21) and ( 22), we find Therefore, the series in (20) converges uniformly on x ∈ (1, M ] for any fixed M > 1 when −x < −1 ⇔ x > 1.Let n be a positive integer.Replacing x by n in (19), we get We split the following summation into two parts.This leads to By utilizing (6), we get On the other hand, we have By using (8), we have Thus, by (11), we have Applying the relation and then making the replacement By combining (23), ( 24), (25), and (26), we have Differentiating both sides of (27) with respect to t yields Since we conclude from (28) that .
By setting t = 1 − s and making the substitution k → k + 1 (and after a short computation), we have . Now, the desired result is obtained by using the simple relation Remark 3.1.By setting t = 1 − s in (27), we get the following elegant alternative representation for log s: Theorem 3.2.Let n ∈ N.For |s − 1| < 1 or |s − 1| = 1 and n ≥ 1, the following identity holds: Proof.Differentiating both sides of (18) with respect to x twice yields We set x = n here and then split the resultant summation into two parts.This gives