A RECURRENCE RELATION FOR BERNOULLI NUMBERS. Mümün Can, Mehmet Cenkci, and Veli Kurt

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1 Bull Korean Math Soc , No 3, pp A RECURRENCE RELATION FOR BERNOULLI NUMBERS Müün Can, Mehet Cenci, and Veli Kurt Abstract In this paper, using Gauss ultiplication forula, a recurrence relation for Bernoulli nubers, generalizing Naias results, is given Introduction Stirling s forula, with any applications in probability theory and statistical physics, is given in the sipler for by n! n n e n, is quite sufficient when n is large Fro a atheatical point of view, the ore accurate expression 2 n! n n e n { 2πn exp 2n 360n n 5 +, is, as is well nown, an exaple of divergent asyptotic series whose perforance gets worse as the nuber of ters is increased beyond a certain value The Bernoulli nubers, denoted by B, are given by the foral power series expansion 3 x e x 0 x B, x < 2π! It is well nown that B 0, B 2, B 2 6,, and B 2+ 0 for > 0 Soe properties and relations for these nubers have been extensively studied by several atheaticians such as Howard[4], Deeba and Rodriguez[3] and Naias[5] Received July 2, Matheatics Subject Classification: Priary B68, 33B5 Key words and phrases: Bernoulli nubers, Stirling s series, Gauss ultiplication forula

2 68 Müün Can, Mehet Cenci, and Veli Kurt In [5], Naias, using Gauss and Gauss-Legendre forulas for the gaa function Γ2n e 2n ΓnΓ π Γ3n 2π 33n 2 ΓnΓ n +, 2 n + Γ 3 n + 2, 3 gave the following recurrence relations for Bernoulli nubers 4 B s B s 2 2 s 3 3 s s 0 s 0 s 2 B, s 2 B + 2 s, respectively He conjectured that an infinite nuber of recurrence relations such as 4 for Bernoulli nubers can be obtained using Gauss ultiplication forula He also conjectured that in all cases the sae Stirling s series is obtained even though the recurrence relation for the coefficients of the series is different In 99, Deeba and Rodriguez[3] presented an eleentary procedure, without use of Gauss ultiplication forula, to obtain an infinite nuber of recurrence relations for the Bernoulli nubers and exhibited distinct recurrence relations for the coefficients of the Stirling series, as conjectured by Naias[5] In particular, using the generating function of Bernoulli nubers, for any positive integer s and any positive integer >, Deeba and Rodriguez[3] obtained 5 B s s s 0 s B j j s, which is a generalization of 4 In 995, Howard[4] proved that 5 is an easy consequence of the ultiplication forula for Bernoulli polynoials In this paper, we show that, sae results can be obtained by using Gauss ultiplication forula, what is exactly conjectured by Naias We also obtain different relations for the coefficients of the Stirling series and in all cases the sae Stirling series is obtained

3 2 Main relation A recurrence relation for Bernoulli nubers 69 For the first step, we note that the gaa function Γz satisfies the Gauss ultiplication forula see [2, p250] 2 Γnz n 2 nz 2π 2 n n i0 Γ z + i n Let, n Z + and > Putting and n instead of n and z in 2 respectively, we get 22 Γn 2 n 2π 2 On the other hand, since Γ n + Γ n + i0 Γ n + i n Γ n n Γ n, 22 becoes 23 Γn2π 2 n 2 Γn i n i Γ n i Since we have Γn n!, Stirling s forula can be written as 24 Γn n n e n 2πn F n, where F n is a function to be deterined Substituting 24 in 23, after necessary operations, we get 25 F n i0 F n i n n e n 2πn n n 2π n e n

4 620 Müün Can, Mehet Cenci, and Veli Kurt n n 2π n n e Since 2π 2 n 2 n n e n 2πn 2 e 2 n+ 2 n n exp + n j ln j n we have, for j, n + 2 j s0 j + j ln x ln j s s s n s j n j s n s j s+ s + s+ n s + { n s j s+ s+ ss + x s s, 2s j s n+ 2 j s+ s s+ n s js 2s s j s 2s s n s Inserting these equations in the right hand-side of the expression 25, we get { s+ + + s+ 26 exp n s ss + s+ s + + s 2s s, after soe anipulation This result iplies that F n can be written as { a s 27 F n exp n s, where the a s are constant coefficients to be deterined Therefore, 24 becoes the series { 28 Γn n n e n a s 2πn exp n s

5 A recurrence relation for Bernoulli nubers 62 Using the left hand-side of the expression 25 and 27, we obtain F n i0 F a s n i exp n s s 29 s j j n We now use the expansion x t x t +, x < t0 Letting x n,, n expression 29, we find a s exp n s s and inserting in the right hand-side of the t0 t a n n t+ In the double suation, let t + s Then we obtain a s s exp n s s a n s s For > s this expression becoes 20 exp n s s a s s j j t+ j j s s a s j j s Equating the coefficients of n in 26 and 20, we obtain the coefficients of Stirling s series as s 2 { a s s+ j s j ss + s 2s + s a j j Notice that for each positive integer, equation 2 gives a recurrence forula for the sequence a s Furtherore, each recurrence forula generates 2, as conjectured by Naias[5] On the other hand, Stirling s forula can be expressed in ters of Bernoulli nubers as see [, p246], 22 Γn n n e n 2πn exp { s2 B s ss n s

6 622 Müün Can, Mehet Cenci, and Veli Kurt Coparing equations 28 and 22, we find 23 a s B s+ ss + Using 2 and 23, we obtain 24 B s+ s+ j s j s 2 s + + s+ j s s + B j s+ 0 j + s +! B + s!! j In 24, writing s instead of s +, we get 5 Relation 24 generalizes both the expressions in 4 using Gauss ultiplication forula, which conjectured by Naias[5] References [] M Abraowitz and I A Stegun, Handboo of Matheatical Functions, National Bureu of Standarts, Appl Math Series [2] T M Apostol, Introduction to Analytic Nuber Theory, Springer-Verlag, 985 [3] E Y Deeba and D M Rodriguez, Stirling s Series and Bernoulli Nubers, Aer Math Monthly 98 99, [4] F T Howard, Applications of a Recurrence for the Bernoulli Nubers, J Nuber Theory , [5] V Naias, A Siple Derivation of Stirling s Asyptotic Series, Aer Math Monthly , Müün Can, Mehet Cenci, and Veli Kurt, Adeniz University, Departent of Matheatics, Antalya, Turey E-ail: can@adenizedutr M Can cenci@adenizedutr M Cenci vurt@adenizedutr V Kurt