Using known zeta-series to derive the Dancs-He series for $,ln{2},$ and $,zeta{(2,n+1)}$

The Riemann zeta function is defined, for real values of s, s > 1, by 1

(1)

For s > 1, the series in Eq. ( 1) converges by the integral test and its sum for integer values of s has attracted much interest since the times of J. Bernoulli, who proved that ∞ k=1 1/k 2 converges to a number between 1 and 2. Further, Euler (1735) proved that this sum evaluates to π 2 /6 , solving the so-called Basel problem. He also studied this kind of series for greater integer values, finding, for even values of s, the notable formula (1750

where n is a positive integer and B 2n are Bernoulli numbers. 2 For odd values of s, s > 1, on the other hand, no analogous closed-form expression is known. In fact, not even an irrationality proof is presently known for ζ(2 n + 1), except for the Apéry proof that ζ(3) is irrational (1978) [2], which makes the things enigmatic. On trying to find out a closed-form expression for ζ(2 n + 1) similar to that in Eq. ( 2), Dancs and He found a new “Euler-type” formula containing series involving the numbers E 2 n+1 (1), where E 2 n+1 (x) denotes the Euler’s polynomial of degree 2 n + 1 [3]. Their main result follows from some intricate series manipulations, in the lines of those found in Tsumura’s proof of Eq. ( 2) [1].

However, the fortuitous appearance of the numbers ln 2 and ζ(2n + 1) in the Dancs-He formulae, which is hard to be explained with usual series expansions, might well remain a mystery. By noting that the numbers E 2 n+1 (1) can be written in terms of B 2n+2 , and then in terms of ζ(2n), via Eq. ( 2), I show here in this work how the Dancs-He series for ln 2 and ζ(2n + 1) can be derived from some known zeta-series.

1 There is also a product representation due to Euler (1749), namely

where the product is taken over all prime numbers p, which is the main reason for the interest of number theorists in this function. As noted by Euler, since the harmonic series diverges, then from Eq. (1) one deduces that lim s→1 + ζ(s) = ∞, which implies, from the product representation, that there is an infinitude of prime numbers.

2 Since B 2n ∈ Q and π is a transcendental number, as first proved by Lindemann (1882), then Eq. ( 2) implies that ζ(2 n) is a transcendental number.

Dancs-He series for ln 2 and ζ(2 n + 1) 3 2. Dancs-He formula for ln 2

For ln 2, Dancs and He found that (see Eq. (2.6) of Ref. [3]):

Theorem 1 (Dancs-He series for ln 2). Let E 2m+1 (x) denote the Euler’s polynomial of degree 2m + 1, m being a nonnegative integer. Then

Proof. Let L be the number at the left-hand side of Eq. ( 1). By noting that

By substituting n = m + 1, one finds that

.

From Euler’s formula for ζ(2n) in Eq. ( 2), one has

.

Now, let us reduce this latter series to a simple closed-form expression.

For this, let us make use of the following series representation for ζ(s) introduced recently by Tyagi and Holm (see Eq. (3.5) in Ref. [4]):

where Γ(x) is the gamma function. 3 As the series in the right-hand side converges when we make s = 1, all we need to do is to take the limit, as s → 1 + , of the factors at the left-hand side. The simple pole of ζ(s) at s = 1 yields lim s→1 + (s -1) • ζ(s) = 1. Also, by applying the l’Hospital rule it is easy to show that lim s→1 1 -2 1-s /(s -1) = ln 2. The product of these two limits yields lim s→1 + ζ(s) 1 -2 1-s = ln 2, thus (5) ln 2

From Eq. ( 3), one has L = ln 2.

Before presenting a general proof for the Dancs-He series for ζ(2n + 1), n being a positive integer, let us tackle the lowest case, i.e. ζ(3), a number for which several series representations have been derived since the times of Euler [6]. For this number, Dancs and He found the following series representation (see Eq. (3.1) of Ref. [3]).

Theorem 2 (Dancs-He series for ζ(3)).

Proof. Let S be the number for which the series at the left-hand side of Eq. ( 2) converges. By substituting E 2m+1 (1) = 2 2 2m+2 -1 2m+2 B 2m+2 in this series, one has

By substituting n = m + 1, one finds that ( 7)

From the relation between B 2n and ζ(2n) in Eq. ( 2), one has (8)

which is valid since the series in Eq. ( 7) converges absolutely. The first series can be easily evaluated from a known summation formula (see Eq. (713) in Ref. [5]), namely

where ζ(s, a) is the Hurwitz (or generalized) zeta function and ζ ′ (s, a) is its derivative with respect to s. 4 As this formula is valid for all t with |t| < 1, it is legitimate to take the limit as t → 1 -on both sides, which yields

The remaining limit is null, since lim

For the second series in Eq. ( 8), let us make use of the Wilton’s formula (Eq. (38) a

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