SERIES REPRESENTATION OF THE SECOND ORDER HYPERGEOMETRIC ZETA FUNCTION HUNDUMA LEGESSE GELETA, SEID MOHAMMED, ABDUL HASSEN Abstract. A Hypergeometric zeta function is a generalization of the Riemann zeta function via integral representation. Hassen and Nguyen in ([2])introduced these families of functions and in subsequent papers, developed many properties analogous to those satisfied by the classical zeta function. (See [2], [3], [4], [5].) They showed that these functions have Dirichlet series type representations with coefficients µ(n, s), which depend on both n and s. In this paper, we will express µ(n, s) explicitly and use this formula to write the hypergeometric zeta function of order 2 as a power series in s with Dirichlet series as coefficients. We also use this series representation to demonstrate that the second order hypergeometric zeta function has a zero free region to the right half plane. 1. Introduction The Riemann zeta function defined by ∞ X 1 ζ(s) = , ns n=1 where s = σ + it, σ > 1, admits an integral representation given by 1 Z ∞ xs−1 ζ(s) = x dx. (1) Γ(s) 0 e − 1 Hassen and Nguyen in [2] introduced and investigated a generalization of the integral representation of the Riemann zeta function by replacing ex − 1 in the denominator in (1) with arbitrary Taylor difference ex − x TN−1(x) where N is a positive integer and TN−1(x) is the Taylor polynomial of e at the origin having degree N −1. This defines a family of what Hassen and Nguyen called hypergeometric zeta functions denoted by ζN (s): 1 Z ∞ xs+N−2 ζN (s) = x dx (2) Γ(s + N − 1) 0 e − TN−1(x) Observe that ζ1(s) = ζ(s). In several papers, ([2], [3]) Hassen and Nguyen established many of the properties of the classical zeta function for the hypergeometric zeta functions. However, the hypergeometric x zeta functions do not appear to have a product formula. The zero of the e − TN−1(x) can be approximated but cannot be found precisely. This makes it difficult to expect a functional equation. On a right half plane, the hypergeometric zeta functions can be represented in the form: ∞ X µN (n, s) ζ (s) = , (3) N ns+N−1 n=1 Date: April, 2013. 2000 Mathematics Subject Classification. Primary 11M41. Key words and phrases. Riemann zeta function, Hypergeometric zeta functions, Coefficients of the hypergeometric zeta functions. 1 2 HUNDUMA LEGESSE GELETA, SEID MOHAMMED, ABDUL HASSEN for σ > 1, where (N−1)(n−1) X ak(N, n)Γ(s + N + k − 1) µ (n, s) = (4) N nkΓ(s + N − 1) k=0 and ak(N, n) is generated by n−1 (N−1)(n−1) x2 x3 xN−1 X (T (x))n−1 = 1 + x + + + ··· + = a (N, n)xk. N−1 2! 3! (N − 1)! k k=0 Following Riemann, it is possible, as Hassen and Nguyen demonstrated, to extend the hypergeometric zeta function to a left half plane. In fact, it has been shown that the hypergeometric zeta functions ζN (s) can be extended analytically to the entire complex plane, except for N simple poles at s = 1, 0, −1, ..., 2 − N. The coefficients µN (n, s) in the series representation of ζN (s) depend on both n and s. It would be desirable to find an expression of µN (n, s) that allows us to write ζN (s) as a linear combination of ordinary Dirichlet series. It is the objective of this paper to investigate the properties of the coefficients µ2(n, s) and write ζ2(s) as a ”power” series in s with Dirichlet Series as coefficients. To this end, we will find an explicit form of the coefficients of these polynomials and rewrite ζ2(s). Γ(s+N+k−1) We note that the expression Γ(s+N−1) which appears in the definition of the coefficient of the hyperge- ometric zeta function of order N as given in(4) is just (s + N − 1)k: Γ(s + N + k − 1) = (s + N − 1) Γ(s + N − 1) k where (s + a)k is Pochhammer symbol: (s + a)k = (s + a)(s + a + 1) ··· (s + a + k − 1), with initial value (s + a)0 = 1. Moreover the following recursive relation holds: (s + a)k+1 = (s + a + k)(s + a)k. To express the polynomials µ2(n, s) explicitly, we need to define a sequence of numbers recursively. For m = 1, 2, ··· , and k = m + 1, m + 2, m + 3, ··· , we define m m m−1 Ak = kAk−1 + Ak−1 , (5) with initial values 0 k Ak = 0,Ak−1 = 1. Note then that k(k + 1) A1 = k! and Ak = 1 + 2 + 3 + ··· + k = . (6) k k 2 We note that the recurrence relation above is the same as that of the unsigned Stirling numbers which are k denoted by and is given by m k k − 1 k − 1 = k + . m m m − 1 Lemma 1.1. (s + 1)k is a polynomial of degree k given as follows: k k k−1 k−1 k−2 k−2 k−3 2 1 (s + 1)k = s + Aks + Ak s + Ak s + ··· + Aks + Ak. SERIES REPRESENTATION OF THE SECOND ORDER HYPERGEOMETRIC ZETA FUNCTION 3 Proof. We shall use induction on k. 1 0 For k = 0 we have (s + 1)0 = 1 = A0s , and assume the assertion is true for k: k k k−1 k−1 k−2 k−2 k−3 2 1 (s + 1)k = s + Aks + Ak s + Ak s + ··· + Aks + Ak, we first observe that the leading coefficient in (s + 1)k+1 is 1. From the Pochhammer symbol since we have, (s + 1)k+1 = (s + 1)k(s + k + 1) k−j k−j−1 the coefficient of s in (s + 1)k+1 is the sum of the coefficient of s in (s + 1)k and k + 1 times k−j k−j−1 the coefficient of s in (s + 1)k . From the induction hypothesis the coefficient of s in (s + 1)k is k−j k−j k+1−j Ak and the coefficient of s in (s + 1)k is Ak . k−j k+1−j k−j k+1−j Therefore, the coefficient of s in (s + 1)k+1 is (k + 1)(Ak ) + Ak = Ak+1 , as desired. Returning to the series representation of the hypergeometric zeta function given in (3), we note that for N = 2 we have, ∞ X µ2(n, s) ζ (s) = , 2 ns+1 n=1 where n−1 X ak(2, n)Γ(s + k + 1) µ (n, s) = 2 nkΓ(s + 1) k=0 and the ak(2, n) is generated by, n−1 X n − 1 (T (x))n−1 = (1 + x)n−1 = xk. 1 k k=0 Thus, n−1 n−1 n−1 n−1 X Γ(s + k + 1) X (s + 1)k µ (n, s) = k = k . (7) 2 nkΓ(s + 1) nk k=0 k=0 Now we want to write these coefficients of the hypergeometric zeta function as a polynomial whose coefficient is explicit as the following lemma shows: Lemma 1.2. µ2(n, s) can be written as a polynomial of degree ”n − 1” with its explicit coefficients as given below: n−1 X m−1 µ2(n, s) = bnms m=1 where n−1 X n − 1 b = n−j Am nm j j j=m−1 Proof. To see this we begin from the very definition of µ2(n, s) n−1 X n − 1 µ (n, s) = n−k (s + 1) . 2 k k k=0 4 HUNDUMA LEGESSE GELETA, SEID MOHAMMED, ABDUL HASSEN m−1 By Lemma 1.1, the coefficients of s in (s + 1)k is n−1 X m Aj j=m−1 for each m = 1, 2, ··· and hence the coefficient of sm−1 in n−1 X n − 1 µ (n, s) = n−k (s + 1) 2 k k k=0 becomes n−1 X n − 1 n−j Am j j j=m−1 which is equal to bnm. As an example we can have the first few µ2(n, s) as follows: µ2(1, s) = 1 = b11 −1 −1 µ2(2, s) = 1 + 2 + 2 s = b21 + b22s 2 2 2 2 2 µ (3, s) = 1 + 3−1 + 2!3−2 + [ 3−1 + 3−2A2]s + 3−2s2 2 1 2 1 2 2 2 2 = b31 + b32s + b33s 3 3 3 3 3 3 µ (4, s) = 1 + 4−1 + 2!4−2 + 3!4−3 + 4−1 + 4−2A2 + 4−3A2 s 2 1 2 3 1 2 2 3 3 3 3 3 + 4−2 + 4−3A3 s2 + 4−3s3 2 3 3 3 2 3 = b41 + b42s + b43s + b44s . Observe that 3 X m−1 µ2(4, s) = b4ms , m=1 where 3 X 3 b = 4−j Am. 4m j j j=m−1 2. Series Representation We are now in a position to state and prove our main result: Theorem 2.1. The hypergeometric zeta function of order 2 can be rewritten as follows ∞ X m−1 ζ2(s) = Dm(s)s , m=1 where the coefficient Dm(s) is a Dirichlet series for each m = 1, 2, 3, ··· . SERIES REPRESENTATION OF THE SECOND ORDER HYPERGEOMETRIC ZETA FUNCTION 5 Proof. We can rewrite ζ2(s) as follows: ∞ X µ2(n, s) ζ (s) = 2 ns+1 n=1 µ (2, s) µ (3, s) µ (4, s) = µ (1, s) + 2 + 2 + 2 + ··· 2 2s+1 3s+1 4s+1 1 + 2−1 + 2−1s 1 + 23−1 + 22!3−2 + [23−1 + 23−2A2]s + 23−2s2 = 1 + + 1 2 1 2 2 2 + ··· 2s+1 3s+1 b b + b s b + b s + b s2 = 11 + 21 22 + 31 32 33 + ··· 1s+1 2s+1 3s+1 b b b = 11 + 21 + 31 + ··· 1s+1 2s+1 3s+1 b b b b b b + 22 + 32 + 42 + ··· s + ··· + nj + (n+1)j + (n+2)j + ··· sj−1 + ··· 2s+1 3s+1 4s+1 ns+1 (n + 1)s+1 (n + 2)s+1 ∞ ∞ ∞ ∞ X bn1 X bn2 X bn3 X bnj = + s + s2 + ··· + sj−1 + ··· ns+1 ns+1 ns+1 ns+1 n=1 n=2 n=3 n=j ∞ −1 ∞ −1 ∞ −1 ∞ −1 X n bn1 X n bn2 X n bn3 X n bnj = + s + s2 + ··· + sj−1 + ··· .