How to get the explicit formula of Bernoulli number using its generating function?

Question

How to get this Bernoulli number explicit formula:
$$B_k=\sum_{n=0}^k\frac{1}{n+1}\sum_{j=0}^{n}(-1)^j\binom nj j^k$$
by using Bernoulli number's generating function:
$$G(k)=\frac{t}{e^t-1}=\sum_{k=0}^{\infty}B_k\frac{t^k}{k!}$$
Thanks for your any kind help.

Answer 1

By writing $j^k$ as $\left.\frac{d^k}{dx^k}e^{jx}\right|_{x=0}$ we have $$\begin{eqnarray*} \sum_{n=0}^{k}\frac{1}{n+1}\sum_{j=0}^{n}(-1)^j\binom{n}{j}j^k &=& \left.\frac{d^k}{dx^k}\sum_{n=0}^{k}\frac{1}{n+1}\sum_{j=0}^{n}\binom{n}{j}(-1)^j e^{jx}\right|_{x=0}\\&=& \left.\frac{d^k}{dx^k}\sum_{n=0}^{k}\frac{(1-e^x)^n}{n+1}\right|_{x=0}\\\\(A)\qquad&=& \left.\frac{d^k}{dx^k}\sum_{n\geq 0}\frac{(1-e^x)^n}{n+1}\right|_{x=0}\\(B)\qquad&=& \left.\frac{d^k}{dx^k}\frac{x}{e^x-1}\right|_{x=0}\qquad\square.\end{eqnarray*} $$ In $(A)$ we exploit the fact that for any $m>k$, the $k$-th derivative of $(1-e^x)^m$ at the origin is simply zero, since $e^x-1=x+\frac{x^2}{2}+\ldots$ In $(B)$ we exploit the fact that $\sum_{n\geq 0}\frac{z^n}{n+1}=\frac{-\log(1-z)}{z}$, then replace $z$ with $1-e^x$.

Answer 2

The explicit formula \begin{equation}\label{Higgins-Gould-B}\tag{1} B_n=\sum_{k=0}^n\frac1{k+1}\sum_{j=0}^k(-1)^j\binom{k}{j}j^n,\quad n\ge0, \end{equation} has a long history, it appeared in the paper

1. H. W. Gould, Explicit formulas for Bernoulli numbers, Amer. Math. Monthly 79 (1972), 44--51; available online at https://doi.org/10.2307/2978125.

It is a special case of the formula (2.5) in the paper

2. J. Higgins, Double series for the Bernoulli and Euler numbers, J. London Math. Soc. $2$nd Ser. 2 (1970), 722--726; Available online at http://dx.doi.org/10.1112/jlms/2.Part_4.722.

Its equivalent form is \begin{equation}\label{Bernoulli-Stirling-eq}\tag{2} B_n=\sum_{k=0}^n(-1)^k\frac{k!}{k+1}S(n,k), \quad n\ge0, \end{equation} where $S(n,k)$ is the Stirling numbers of the second kind. See page 29, Remark 2 in the paper

3. Bai-Ni Guo and Feng Qi, An explicit formula for Bernoulli numbers in terms of Stirling numbers of the second kind, Journal of Analysis & Number Theory 3 (2015), no. 1, 27--30.

There existed at least seven alternative proofs of the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} in the paper [1, 2] above and in the following monographs and papers:

4. L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Revised and Enlarged Edition, D. Reidel Publishing Co., Dordrecht and Boston, 1974, page 220.
5. R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics---A Foundation for Computer Science, 2nd ed., Addison-Wesley Publishing Company, Reading, MA, 1994.
6. B.-N. Guo and F. Qi, Alternative proofs of a formula for Bernoulli numbers in terms of Stirling numbers, Analysis (Berlin) 34 (2014), no. 2, 187--193; available online at http://dx.doi.org/10.1515/anly-2012-1238.
7. F. Qi and B.-N. Guo, Alternative proofs of a formula for Bernoulli numbers in terms of Stirling numbers, Analysis (Berlin) 34 (2014), no. 3, 311--317; available online at http://dx.doi.org/10.1515/anly-2014-0003.

To the best of my knowledge, except the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} above, there are also the following explicit formulas for the Bernoulli numbers $B_n$: \begin{align}\label{Higgins-Gould-B(11)}\tag{3} B_n&=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{n!}{(n+j)!}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^{n+j}, \quad n\ge0;\\ B_n&=\sum_{i=0}^n(-1)^{i}\frac{\binom{n+1}{i+1}}{\binom{n+i}{i}}S(n+i,i), \quad n\ge0;\label{Bernoulli-Stirling-formula}\tag{4}\\ B_{2k}&=1+\sum_{m=1}^{2k-1}\frac{S(2k+1,m+1) S(2k,2k-m)}{\binom{2k}{m}}\\ &\quad-\frac{2k}{2k+1}\sum_{m=1}^{2k}\frac{S(2k,m)S(2k+1,2k-m+1)}{\binom{2k}{m-1}}, \quad k\in\mathbb{N};\tag{5}\\ B_{2k}&=\frac{(-1)^{k-1}k}{2^{2(k-1)}(2^{2k}-1)}\sum_{i=0}^{k-1}\sum_{\ell=0}^{k-i-1} (-1)^{i+\ell}\binom{2k}{\ell}(k-i-\ell)^{2k-1}, \quad k\in\mathbb{N};\tag{6}\\ B_{2m}&=(-1)^{m-1}\frac{m} {2^{2m-1}\bigl(2^{2m}-1\bigr)}\Biggl[\sum_{k=0}^{m-1} (-1)^k\binom{2m}{k}(m-k)^{2m-1}\\ &\quad+2\sum_{k=1}^{m-1}(-1)^k\sum_{\ell=0}^{m-k-1} (-1)^{\ell}\binom{2m}{\ell}(m-k-\ell)^{2m-1}\Biggr],\quad m\in\mathbb{N};\\ B_{2m}&=\frac{m} {2^{2m-1}\bigl(2^{2m}-1\bigr)}\sum_{\ell=1}^{2m}\frac{(-1)^{\ell-1}}{2^\ell} \biggl(\frac1\ell-\frac1{m+1}\biggr) \binom{2m+1}{\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{2m}, \quad m\in\mathbb{N};\\ B_{2k}&= \frac12 - \frac1{2k+1} - 2k \sum_{i=1}^{k-1} \frac{A_{2(k-i)}}{2(k - i) + 1},\quad k\in\mathbb{N};\tag{7} \end{align} where $A_m$ is defined by \begin{equation*} \sum_{m=1}^nm^k=\sum_{m=0}^{k+1}A_mn^{m}. \end{equation*} The formulas \eqref{Higgins-Gould-B(11)} and \eqref{Bernoulli-Stirling-formula} are also equivalent to eah other.

By the way, I would like to mention two intereting double inequalities related to the Bernoulli numbers $B_{2n}$ as follows.

• The double inequality \begin{equation}\label{Bernoulli-ineq}\tag{8} \frac{2(2n)!}{(2\pi)^{2n}} \frac{1}{1-2^{\alpha -2n}} \le |B_{2n}| \le \frac{2(2n)!}{(2\pi)^{2n}}\frac{1}{1-2^{\beta -2n}} \end{equation} is valid for $n\in\mathbb{N}$, where $\alpha=0$ and $ \beta=2+\frac{\ln(1-6/\pi^2)}{\ln2}=0.6491\dotsc $ are the best possible in the sense that they can not be replaced respectively by any bigger and smaller constants in the double inequality \eqref{Bernoulli-ineq}. See the paper [8] below.
• The ratios $\frac{|B_{2(n+1)}|}{|B_{2n}|}$ for $n\in\mathbb{N}$ can be bounded by \begin{equation}\label{ineq-Bernou-equiv}\tag{9} \frac{2^{2n-1}-1}{2^{2n+1}-1}\frac{(2n+1)(2n+2)}{\pi^2} <\frac{|B_{2(n+1)}|}{|B_{2n}|} <\frac{2^{2n}-1}{2^{2n+2}-1}\frac{(2n+1)(2n+2)}{\pi^2}. \end{equation} See the paper [9] below.

More related references

8. H. Alzer, Sharp bounds for the Bernoulli numbers, Arch. Math. (Basel) 74 (2000), no. 3, 207--211; available online at https://doi.org/10.1007/s000130050432.
9. Feng Qi, A double inequality for the ratio of two non-zero neighbouring Bernoulli numbers, Journal of Computational and Applied Mathematics 351 (2019), 1--5; available online at https://doi.org/10.1016/j.cam.2018.10.049.
10. Sumit Kumar Jha, Two new explicit formulas for the Bernoulli numbers, Integers 20 (2020), Paper No. A21, 5 pp.
11. Sumit Kumar Jha, Two new explicit formulas for the even-indexed Bernoulli numbers, J. Integer Seq. 23 (2020), no. 2, Art. 20.2.6, 6 pp.
12. Sumit Kumar Jha, A new explicit formula for Bernoulli numbers involving the Euler number, Mosc. J. Comb. Number Theory 8 (2019), no. 4, 385--387; availble online at https://doi.org/10.2140/moscow.2019.8.389.
13. B.-N. Guo and F. Qi, Some identities and an explicit formula for Bernoulli and Stirling numbers, J. Comput. Appl. Math. 255 (2014), 568--579; available online at http://dx.doi.org/10.1016/j.cam.2013.06.020.
14. S.-L. Guo and F. Qi, Recursion formulae for $\sum_{m=1}^nm^k$, Z. Anal. Anwendungen 18 (1999), no. 4, 1123--1130; available online at http://dx.doi.org/10.4171/ZAA/933.
15. S. Jeong, M.-S. Kim, and J.-W. Son, On explicit formulae for Bernoulli numbers and their counterparts in positive characteristic, J. Number Theory 113 (2005), no. 1, 53; available online at http://dx.doi.org/10.1016/j.jnt.2004.08.013.
16. F. Qi, Derivatives of tangent function and tangent numbers, Applied Mathematics and Computation 268 (2015), 844--858; available online at https://doi.org/10.1016/j.amc.2015.06.123.
17. L. Saalschutz, Vorlesungen uber die Bernoulli'schen Zahlen, ihren Zusammenhang mit den Secanten-Coefficienten und ihre wichtigeren Anwendungen, Berlin, 1893. Available since 1964 in Xerographed form from University Microfilms, Ann Arbor, Michigan. Order No. OP-17136.
18. S. Shirai and K.-I. Sato, Some identities involving Bernoulli and Stirling numbers, J. Number Theory 90 (2001), no. 1, 130--142; available online at http://dx.doi.org/10.1006/jnth.2001.2659.
19. Chao-Ping Chen and Feng Qi, Three improper integrals relating to the generating function of Bernoulli numbers, Octogon Mathematical Magazine 11(2003), no. 2, 408--409.
20. Ye Shuang, Bai-Ni Guo, and Feng Qi, Logarithmic convexity and increasing property of the Bernoulli numbers and their ratios, Revista de la Real Academia de Ciencias Exactas Fisicas y Naturales Serie A Matematicas 115 (2021), no. 3, Paper No. 135, 12 pages; available online at https://doi.org/10.1007/s13398-021-01071-x.
21. Feng Qi, Notes on a double inequality for ratios of any two neighbouring non-zero Bernoulli numbers, Turkish Journal of Analysis and Number Theory 6 (2018), no. 5, 129--131; available online at https://doi.org/10.12691/tjant-6-5-1.
22. Z.-H. Yang and J.-F. Tian, Sharp bounds for the ratio of two zeta functions, J. Comput. Appl. Math. 364 (2020), 112359, 14 pages; available online at https://doi.org/10.1016/j.cam.2019.112359.
23. L. Zhu, New bounds for the ratio of two adjacent even-indexed Bernoulli numbers, Rev. R. Acad. Cienc. Exactas Fis. Nat. Ser. A Mat. RACSAM 114 (2020), no. 2, Paper No. 83, 13 pages; available online at https://doi.org/10.1007/s13398-020-00814-6.

Answer 3

In the paper [2], the Bernoulli polynomials $B_n(x)$ were determinantally expressed as \begin{equation} B_n(x)=\frac{(-1)^n}{(n-1)!} \begin{vmatrix} 1 & x & x^2 & x^3 & \dotsm & x^{n-1} & x^n\\ 1 & \frac12 & \frac13 & \frac14 & \dotsm & \frac1n & \frac1{n+1}\\ 0 & 1 & 1 & 1 & \dotsm & 1 & 1\\ 0 & 0 & 2 & 3 & \dotsm & n-1 & n\\ 0 & 0 & 0 & \binom32 & \dotsm & \binom{n-1}2 & \binom{n}2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \dotsm & \binom{n-1}{n-2} & \binom{n}{n-2} \end{vmatrix}, \quad n\in\mathbb{N}. \end{equation} In the paper [1] below, the Bernoulli polynomials $B_n(x)$ were represented as \begin{align*} B_n(x)&=(-1)^nn! \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ \frac{x}{1!} & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ \frac{x^2}{2!} & \frac1{3!} & \frac1{2!} & 1 & 0 & 0 & \dotsm & 0\\ \frac{x^3}{3!} & \frac1{4!} & \frac1{3!} & \frac1{2!} & 1 & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ \frac{x^n}{n!} & \frac1{(n+1)!} & \frac1{n!} & \frac1{(n-1)!} & \frac1{(n-2)!}&\frac1{(n-3)!}&\dotsm& 1 \end{vmatrix}\\ &=(-1)^n\frac{n!}{\prod_{k=1}^nk!} \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ x & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ x^2 & \frac{2!}{3!} & 1 & 2! & 0 & 0 & \dotsm & 0\\ x^3 & \frac{3!}{4!} & 1 & \frac{3!}{2!} & 3! & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ x^n & \frac{n!}{(n+1)!} & 1 & \frac{n!}{(n-1)!} & \frac{n!}{(n-2)!}&\frac{n!}{(n-3)!}&\dotsm& n! \end{vmatrix}\\ &=(-1)^n\prod_{k=1}^{n-1}\frac{(k-1)!}{k!} \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ x & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ x^2 & \frac{2!}{3!} & 1 & 2! & 0 & 0 & \dotsm & 0\\ x^3 & \frac{3!}{4!} & 1 & \frac{3!}{2!} & \frac{3!}{2!} & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ x^n & \frac{n!}{(n+1)!} & 1 & \frac{n!}{(n-1)!} & \frac{n!}{(n-2)!2!}&\frac{n!}{(n-3)!3!}&\dotsm& \frac{n!}{2!(n-2)!} \end{vmatrix}. \end{align*} In the paper [4], the Bernoulli polynomials $B_n(x)$ was explicitly expressed as \begin{multline}\label{BP-Stirl-form} B_n(x)=\sum_{k=1}^nk!\sum_{r+s=k}\sum_{\ell+m=n}(-1)^m\binom{n}{\ell} \frac{\ell!}{(\ell+r)!}\frac{m!}{(m+s)!}\\ \times\Biggl[\sum_{i=0}^r\sum_{j=0}^s(-1)^{i+j}\binom{\ell+r}{r-i}\binom{m+s}{s-j}S(\ell+i,i)S(m+j,j)\Biggr]x^{m+s}(1-x)^{\ell+r} \end{multline} and was determinantally represented as \begin{equation}\label{Bern-Polyn-determ} B_n(x)=(-1)^n\biggl|\frac1{\ell+1}\binom{\ell+1}{m} \bigl[(1-x)^{\ell-m+1}-(-x)^{\ell-m+1}\bigr]\biggr|_{1\le \ell\le n,0\le m\le n-1} \end{equation} for $n\in\mathbb{N}$, where $S(n,k)$ denotes the Stirling numbers of the second kind and $|\cdot|_{1\le \ell\le n,0\le m\le n-1}$ denotes a $n\times n$ determinant.

In the paper [5], an alternative determinantal expression \begin{equation}\label{Bernoulli-Polyn-Det-Erew} B_n(x)=\frac{(-1)^n}{(n+1)!} \begin{vmatrix} 1&1&0&0&\dotsm&0&0&0\\ \frac{x}2&\binom{2}0&\frac12&0&\dotsm&0&0&0\\ \frac{x^2}3&\binom{3}0&\binom{3}1&\frac13&\dotsm&0&0&0\\ \frac{x^3}4&\binom{4}0&\binom{4}1&\binom{4}2&\dotsm&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ \frac{x^{n-2}}{n-1}&\binom{n-1}0&\binom{n-1}1&\binom{n-1}2&\dotsm &\binom{n-1}{n-3}&\frac1{n-1}&0\\ \frac{x^{n-1}}{n}&\binom{n}0&\binom{n}1&\binom{n}2&\dotsm &\binom{n}{n-3}&\binom{n}{n-2}&\frac1{n}\\ \frac{x^n}{n+1}&\binom{n+1}0&\binom{n+1}1&\binom{n+1}2&\dotsm &\binom{n+1}{n-3}&\binom{n+1}{n-2}&\binom{n+1}{n-1} \end{vmatrix}, \quad n\ge0 \end{equation} for the Bernoulli numbers $B_n(x)$ was derived.

Theorem 1.1 in [3] reads that, for all integers $n,r\ge0$, the Bernoulli polynomials $B_n(r)$ can be computed in terms of the $r$-Stirling numbers of the second kind $S_r(n,k)$ by \begin{equation}\label{Bernoulli-Poly-r-Stirling-eq} B_n(r)=\sum_{k=0}^n(-1)^k\frac{k!}{k+1}S_r(n,k). \end{equation}

When taking $x=0$ in the above determinatal expressions and closed-form expressions, we can arrive at closed-form formulas for the Bernoulli numbers $B_n=B_n(0)$.

References

1. R. Booth and H. D. Nguyen, Bernoulli polynomials and Pascal's square, Fibonacci Quart. 46/47 (2008/2009), no. 1, 38--47.
2. F. Costabile, F. Dell'Accio, M. I. Gualtieri, A new approach to Bernoulli polynomials, Rend. Mat. Appl. (7) 26 (2006), no. 1, 1--12.
3. B.-N. Guo, I. Mezo, and F. Qi, An explicit formula for the Bernoulli polynomials in terms of the $r$-Stirling numbers of the second kind, Rocky Mountain J. Math. 46 (2016), no. 6, 1919--1923; available online at https://doi.org/10.1216/RMJ-2016-46-6-1919.
4. F. Qi and R. J. Chapman, Two closed forms for the Bernoulli polynomials, J. Number Theory 159 (2016), 89--100; available online at https://doi.org/10.1016/j.jnt.2015.07.021.
5. F. Qi and B.-N. Guo, Some determinantal expressions and recurrence relations of the Bernoulli polynomials, Mathematics 4 (2016), no. 4, Article 65, 11 pages; available online at https://doi.org/10.3390/math4040065.