Explicit Formulas for Some Infinite 3F2(1)-Series

We establish two recurrence relations for some Clausen’s hypergeometric functions with unit argument. We solve them to give the explicit formulas. Additionally, we use the moments of Ramanujan’s generalized elliptic integrals to obtain these recurrence relations.

The case p+1 F p when p = 1 is called the Gauss hypergeometric function. The following well-known and celebrated summation formula for 2 F 1 (1) is due to Gauss: Another interesting formula for 3 F 2 (1) is due to Ramanujan: where n is a positive integer and which is obtained by replacing n by n − 1 in Entry 29(b) in ( [2], p. 39). This formula was stated without proof by Ramanujan in his first letter to Hardy. There are numerous hypergeometric series identities in mathematical literature (see [3,4]). The evaluation of the hypergeometric sum 3 F 2 (1) (the Clausenian hypergeometric function with unit argument) is of ongoing interest, since it appears ubiquitously in many physics and statistics problems [5][6][7]. The q-extension of the 3F2(1)series is also very interesting and has been studied by many researchers; for example, see [8] and the references therein.
Recently, Asakura et al. [9] proved that where B(a, b) = Γ(a)Γ(b)/Γ(a + b) is the beta function and the right hand side denotes the Q-linear subspace of C generated by 1, 2πi, and log(α)'s, α ∈ Q × under some conditions on a, b, q ∈ Q\Z.
Thus, our hypergeometric series F(x) can be got by setting s = 1/3, and we have In the last section, we will use the moments of Ramanujan's generalized elliptic integral to give another method of obtaining the explicit evaluations.
The organization of this paper is as follows. In Section 2, we give some preliminaries. We provide two recurrence relations for the hypergeometric series F(x). Then, we solve these recurrence relations to obtain explicit evaluations of the hypergeometric series F(n ± q) for n ∈ Z and q = 1, 1/2, 1/3, 2/3, 1/4, 3/4 in Section 3. In Section 4, we list the explicit forms of F(n ± q) for n = ±1, ±2, ±3. In the final section, we use the moments of Ramanujan's generalized elliptic integral to give another method of obtaining the same evaluations.

Preliminaries
We list an explicit formula in ( [12], Equation 3.13-(41)) which we need to use later.
We first prove a useful lemma.
Lemma 1. Let x be a complex number with x / ∈ Z ≤0 . Then Proof. We rewrite this hypergeometric series F(x) as .
We use the partial fraction decomposition of (n + 1 6 )(n + 5 to the above identity, we have The first two hypergeometric series in the right-hand side of the above equation can be evaluated by Equation (7) and the Gauss formula Equation (1) Substituting these values into the last equation of our 3 F 2 (1), we can get the required formula.
We reverse Equation (8) and get another recurrence relation for our 3 F 2 (1). Lemma 2. Let x be a complex number with x / ∈ Z ≤0 and x = 1 6 , 5 6 . Then We have listed the explicit formulas of , and D = arccos Let ζ = e 2πi/5 , ζ 20 = e 2πi/20 , α = 1/ 10 √ 24 > 0, where log(x) takes the principal values, This formula for q = k/5 is complicated. It can be seen that, although the results in Theorems 1 and 2 can be used to obtain their general formulas, the formulas will be more cumbersome and complicated, so we will not deal with the formula and its general form for q = k/5 in this paper.
We give examples applying Equation (8) where A, B are defined in Equation (6) and C, D are defined in Equation (10).

Explicit Formulas
We will solve the recurrence relations in Lemmas 1 and 2 as explicit formulas in this section. Theorem 1. Let m be a non-negative integer, p be a non-zero complex number such that m + p + 1 is not a non-positive integer. Then Proof. Consider the hypergeometric series F(x) with x > 1. So we can decomposite x = m + p, where m ∈ N. Applying the recurrence relation in Lemma 2 a positive integer times, we get 3 F 2 1 6 , 5 6 , x 1, .
Using the mathematical induction on the integer , it is easy to prove that the above formula is correct. We use the Pochhammer symbols to rewrite the function T and note that x = m + p, we have Therefore, 3 F 2 1 6 , 5 6 , m + p 1, m + p + 1 1 = T(m) 3 F 2 1 6 , 5 6 , p 1, p + 1 1 + 1 2π Our result is followed by the fact p(p + 1) m = (p) m (p + m).
Followed by using the similar method to the recurrence relation in Lemma 1, we have the explicit formula for F(p − m): Furthermore, we use the Pochhammer symbols at the negative integer index (a) −k , which is defined by Then, the explicit formula F(p − m) is symmetry to the formula F(p + m).

Examples
In this section, we list some concrete examples of our results by using Equations (17)-(26) with n = 1, 2. First we indicate that our formula also cover the most well-known formula, for n ∈ N, with the parameters p = 1 and m = n − 1 ∈ N 0 in Theorem 1. This identity can be derived by using the more general formula about 3 F 2 a, b, n c, n + p 1 in ( [13], Equation (1.7)), or ( [3], Equation (16)), where n, p are positive integers. It is note that we use the notations A, B defined in Equation (6) and C, D defined in Equation (10)    Thus, our hypergeometric series F(x) can be got by setting s = 1/3, and we have The following formula is in ( [11], Equation (29)).
This is exactly the same recurrence relation Equation (8)