Two New Explicit Formulas for the Even-Indexed Bernoulli Numbers

Sumit Kumar Jha International Institute of Information Technology Hyderabad-500 032 India

              kumarjha.sumit@research.iiit.ac.in

Abstract We give two new explicit formulas for the even-indexed Bernoulli numbers in terms

of the Stirling numbers of the second kind.

1 Introduction

Definition 1. The Bernoulli numbers B_{n} can be defined by the following generating function:

\frac{t}{e^{t} − 1} = ∑_{n≥0} \frac{B_{n}t^{n}}{n!} ,

where |t| < 2π.

Definition 2. The Stirling number of the second kind, denoted by S(n, m), is the number of ways of partitioning a set of n elements into m nonempty sets.

There are many explicit formulas known for the Bernoulli numbers [1, 9, 4, 5, 6]. For example, all of the formulas below express the Bernoulli numbers explicitly in terms of the Stirling numbers of the second kind:

B_{r} = ∑_{k=1}^{r}(−1)^{k} · k! · \frac{S(r, k)}{k + 1},

B_{r} = ∑_{k=1}^{r}(−1)^{k−1} · (k − 1)! · \frac{S(r, k)}{k + 1},

B_{r+1} = ∑_{k=1}^{r} \frac{(−1)^{k−1} k! S(r, k)}{(k + 1)(k + 2)},

B_{r} = \frac{r}{1 − 2^{r}} ∑_{k=1}^{r−1} \frac{(−1)^{k} k! S(r − 1, k)}{2^{k+1}},

B_{r+1} = \frac{(−1)^{r} · (r + 1) · 2^{r−1}}{2^{r+1} − 1} ∑_{k=1}^{r} \frac{(−1)^{k} k! S(r, k)}{k + 1} · 2^{−2k}\binom{2k}{k},

B_{r} = ∑_{i=0}^{r}(−1)^{i} \begin{matrix}( & {r+1 \atop i+1} & ) \\ ( & {r+i \atop i} & ) \end{matrix}S(r + i, i),

B_{r+1} = −\frac{r + 1}{4(1 + 2^{−(r+1)}(1 − 2^{−r}))} (∑_{k=1}^{r}(−1)^{k} · \frac{S(r, k)}{k + 1} · (\frac{3}{4})^{(k)} + 4^{−r}E_{r}) ,  (1)

where (E_{r}) denotes the Euler numbers defined by the following generating function:

\frac{1}{cosh t} = \frac{2}{e^{t} + e^{−t}} = ∑_{n=0}^{∞} \frac{E_{n}}{n!} · t^{n}.

In the following section, we derive two new explicit formulas for the even-indexed Bernoulli numbers in terms of the Stirling numbers of the second kind.

2 Main results

Our main results are the following.

Theorem 3. For all positive integers r we have

B_{2r} = \frac{−4r}{3(3 − 3^{1−2r})} ∑_{k=1}^{2r−1}(−1)^{k} \frac{S(2r − 1, k)}{k + 1} (\frac{2}{3})^{(k)} ,  (2)

and

B_{2r} = \frac{2r}{3((1 − 2^{1−2r})(1 − 3^{1−2r}) − 2)} ∑_{k=1}^{2r−1}(−1)^{k} \frac{S(2r − 1, k)}{k + 1} (\frac{5}{6})^{(k)} ,  (3)

where S(2r − 1, k) denotes the Stirling numbers of the second kind, and

x^{(n)} = x(x + 1)(x + 2) ⋯ (x + n − 1)

denotes the rising factorial.

To prove the above result, we first recall the following fact Theorem 4. [10] The nth Bernoulli polynomial, B_{n}(t), defined by

B_{n}(t) = ∑_{j=0}^{n} \binom{n}{j}B_{n−j}t^{j}

takes the following ‘special’ values at certain rational numbers with small denominators B_{n}(1) = B_{n}(0) = B_{n} for n ≥ 1,

B_{n} (\frac{1}{2}) = (2^{1−n} − 1)B_{n},

B_{2n} (\frac{1}{4}) = B_{2n} (\frac{3}{4}) = \frac{2 − 2^{2n}}{4^{2n}} B_{2n},  (4)
B_{2n} (\frac{1}{3}) = B_{2n} (\frac{2}{3}) = \frac{3 − 3^{2n}}{2 · 3^{2n}} B_{2n},  (5)
B_{2n} (\frac{1}{6}) = B_{2n} (\frac{5}{6}) = \frac{(2 − 2^{2n})(3 − 3^{2n})}{2 · 6^{2n}}B_{2n}.  (6)

Remark 5. Granville and Sun [2] noted that, “It is not known if B_{n}(a/q) has as simple a ‘closed form’ for any other rational a/q with 1 ≤ a ≤ q − 1 and (a, q) = 1, though this has long been considered an interesting question.”

Our proof also requires the following result. Lemma 6. For all 0 < t < 1 and l ∈ ℕ we have

∫_{0}^{∞} x^{t−1} \frac{Li_{−l}(−x)}{1 + x} dx = \frac{π}{sin tπ} (\frac{B_{l+1}(1 − t) − B_{l+1}}{l + 1} ) ,  (7)

where Li_{−l}(−x) is the negative polylogarithm function, and B_{r}(1 − t) denotes the Bernoulli polynomial. Proof. Consider the following generating function [8]

\frac{Li_{−r}(−x)}{1 + x} = ∑_{n=0}^{∞} S_{r}(n)(−x)^{n},  (8)

where S_{r}(n) = ∑_{k=1}^{n} k^{r} for n ≥ 1, and S_{r}(0) = 0.

We use Ramanujan’s master theorem (RMT) from [3] that states that

∫_{0}^{∞} x^{t−1}\{ϕ(0) − xϕ(1) + x^{2}ϕ(2) − ⋯ \} dx = \frac{π}{sin tπ} ϕ(−t),

where the integral is convergent for 0 < ℜ(t) < 1, and after certain conditions are satisfied by ϕ.

Using RMT with Eq. (8) gives us Eq. (7).

Proof of Theorem 3. Substituting t = 2/3 and l = 2r − 1 in Eq. (7) gives us

∫_{0}^{∞} x^{−1/3} \frac{Li_{1−2r}(−x)}{1 + x} dx = \frac{2π}{\sqrt{3}} (\frac{B_{2r}(1/3) − B_{2r}}{2r} ) .

We use the following representation from the note [7]

Li_{1−2r}(−x) = ∑_{k=1}^{2r−1} k!S(2r − 1, k) (\frac{1}{1 + x})^{k+1} (−x)^{k}  (9)

to conclude that

∫_{0}^{∞} x^{−1/3} \frac{Li_{1−2r}(−x)}{1 + x} dx = ∑_{k=1}^{2r−1}(−1)^{k} · k! S(2r − 1, k) ∫_{0}^{∞} \frac{x^{k−1/3}}{(1 + x)^{k+2}} dx

= ∑_{k=1}^{2r−1}(−1)^{k} · k! S(2r − 1, k)\frac{Γ(k + 2/3)Γ(4/3)}{Γ(k + 2)}

= ∑_{k=1}^{2r−1}(−1)^{k} · \frac{S(2r − 1, k)}{k + 1} (\frac{2}{3})^{(k)} · \frac{2π}{3\sqrt{3}},

where Γ(·) denotes the Gamma function. Recalling Eq. (5) we have

B_{2r}(1/3) − B_{2r} = − (3 − 3^{1−2r}) \frac{B_{2r}}{2} .

Now we can readily conclude Eq. (2).

To prove Eq. (3) we substitute t = 5/6 and l = 2r − 1 in the Eq. (7) to get

∫_{0}^{∞} x^{−1/6} \frac{Li_{1−2r}(−x)}{1 + x} dx = 2π (\frac{B_{2r}(1/6) − B_{2r}}{2r} ) .

The representation (9) also lets us to conclude that

∫_{0}^{∞} x^{−1/6} \frac{Li_{1−2r}(−x)}{1 + x} dx = ∑_{k=1}^{2r−1}(−1)^{k} · k! S(2r − 1, k) ∫_{0}^{∞} \frac{x^{k−1/6}}{(1 + x)^{k+2}} dx

= ∑_{k=1}^{2r−1}(−1)^{k} · k! S(2r − 1, k)\frac{Γ(k + 5/6)Γ(7/6)}{Γ(k + 2)}

= ∑_{k=1}^{2r−1}(−1)^{k} · \frac{S(2r − 1, k)}{k + 1} (\frac{5}{6})^{(k)} · \frac{π}{3} .

Recalling Eq. (6) we have

B_{2r}(1/6) − B_{2r} = \frac{B_{2r}}{2} ((1 − 2^{1−2r})(1 − 3^{1−2r}) − 2).

Now we can readily conclude Eq. (3).

Remark 7. Substituting t = 3/4 and l = 2r − 1 in Eq. (7), and using Eq. (4) we can obtain

B_{2r} = −\frac{2r}{4(1 + 2^{−2r}(1 − 2^{−(2r−1)}))} ∑_{k=1}^{2r−1}(−1)^{k} · \frac{S(2r − 1, k)}{k + 1} · (\frac{3}{4})^{(k)} .

The above is just a special case of Eq. (1) which was obtained in [6].

References

[1] H. W. Gould, Explicit formulas for Bernoulli numbers, Amer. Math. Monthly 79 (1972), 44–51.

[2] A. Granville and Z. W. Sun, Values of Bernoulli polynomials, Pacific J. Math. 172 (1996), 117–137.

[3] G. H. Hardy, Ramanujan. Twelve Lectures on Subjects Suggested by His Life and Work, 3rd edition, Chelsea, 1978.

[4] S. K. Jha, Two new explicit formulas for the Bernoulli numbers, preprint, 2019. Avail-

   able at https://arxiv.org/abs/1905.11216.

[5] S. K. Jha, A new explicit formula for the Bernoulli numbers in terms of the Stirling numbers of the second kind, preprint, 2019. Available at https://osf.io/95gkw.

[6] S. K. Jha, A new explicit formula for Bernoulli numbers involving the Euler number, Mosc. J. Comb. Number Theory 8 (2019), 385–387.

[7] S. E. Landsburg, Stirling numbers and polylogarithms, preprint, 2009. Available at

    http://www.landsburg.com/query.pdf.

[8] L. Lewin, Polylogarithms and Associated Functions, North-Holland, 1981.

[9] J. Quaintance and H. W. Gould, Combinatorial Identities For Stirling Numbers: The Unpublished Notes Of H. W. Gould, World Scientific, 2015.

[10] Z. H. Sun, Congruences involving Bernoulli and Euler numbers, J. Number Theory 128 (2008), 280–312.

2010 Mathematics Subject Classification: Primary 11B68. Keywords: Bernoulli number, Stirling number of the second kind, polylogarithm function, Ramanujan’s master theorem.

(Concerned with sequences \underline{A000367}, \underline{A002445}, \underline{A008277}, \underline{A027641}, and \underline{A027642}.)

Received November 6 2019; revised versions received November 7 2019; January 19 2020; January 20 2020; January 23 2020; January 27 2020; February 10 2020; February 20 2020. Published in Journal of Integer Sequences, February 20 2020.

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