Is the ratio of consecutive Bernoulli polynomials uniformly bounded

Question

When investigating a certain kind of Stirling's approximation of the Gamma function error terms occur such as \begin{equation} E(s)=\frac{1}{s}\sum_{j=1}^\infty B_{j+1}(a)\frac{(-1)^{j+1}}{j(j+1)s^{j-1}},|s|\to\infty \end{equation} where $a,s\in\mathbb{C}$ and the $B_j$ are the Bernoulli polynomials. I'd like to survey the convergence properties of the sum consideraring the ratio \begin{equation} \left|\frac{B_{j+1}(a)}{B_j(a)}\frac{j-1}{j+1}\frac1s\right|. \end{equation} So the first question is whether $B_{j+1}(a)/B_j(a)$ is bounded for $j\to\infty$ and fixed $a$. This would yield $E(s)=O(1/|s|),|s|\to\infty$.

Now let $a\in K\subset\mathbb{C}$ with $K$ a compact set. Is $B_{j+1}(a)/B_j(a)$ bounded for $j\to\infty$ uniformly in $a\in K$?

Or maybe someone proposes another approach to get \begin{equation} E(s)=O(1/|s|),|s|\to\infty\text{ uniformly in $K$} \end{equation} $K=\{a\}$ resp. $K$ compact.

Answer 1

The Bernoulli numbers $B_n$ can be generated by \begin{equation*} \frac{z}{e^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}=1-\frac{z}2+\sum_{k=1}^\infty B_{2k}\frac{z^{2k}}{(2k)!}, \quad \vert z\vert<2\pi. \end{equation*} Because the function $\frac{x}{e^x-1}-1+\frac{x}2$ is even in $x\in\mathbb{R}$, all of the Bernoulli numbers $B_{2k+1}$ for $k\in\mathbb{N}=\{1,2,3,\dotsc\}$ equal $0$.

The Bernoulli polynomials $B_n(u)$ can be generated by \begin{equation*} \frac{ze^{uz}}{e^z-1}=\sum_{k=0}^{\infty}B_{k}(u) \frac{z^k}{k!},\quad |z|<2\pi. \end{equation*} It is clear that $B_k(0)=B_k$ for $k\in\mathbb{N}_0=\{0,1,2,\dotsc\}$.

In the paper [1] below, Qi obtained a double inequality \begin{equation}\label{ineq-Bernou-equiv} \frac{2^{2k-1}-1}{2^{2k+1}-1}\frac{(2k+1)(2k+2)}{\pi^2} <\frac{|B_{2k+2}|}{|B_{2k}|} <\frac{2^{2k}-1}{2^{2k+2}-1}\frac{(2k+1)(2k+2)}{\pi^2}, \quad k\in\mathbb{N}.\tag{1} \end{equation} I think this double inequality is an answer to this question.

The double inequality \eqref{ineq-Bernou-equiv} for the ratio of two non-zero neighbouring Bernoulli numbers $B_{2k}$ has been further investigated in the papers [2, 3, 4, 5] below.

References

1. F. Qi, A double inequality for the ratio of two non-zero neighbouring Bernoulli numbers, J. Comput. Appl. Math. 351 (2019), 1--5; available online at https://doi.org/10.1016/j.cam.2018.10.049.
2. F. Qi, Notes on a double inequality for ratios of any two neighbouring non-zero Bernoulli numbers, Turkish J. Anal. Number Theory 6 (2018), no. 5, 129--131; available online at https://doi.org/10.12691/tjant-6-5-1.
3. Y. Shuang, B.-N. Guo, and F. Qi, Logarithmic convexity and increasing property of the Bernoulli numbers and their ratios, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM 115 (2021), no. 3, Paper No. 135, 12 pages; available online at https://doi.org/10.1007/s13398-021-01071-x.
4. 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.
5. L. Zhu, New bounds for the ratio of two adjacent even-indexed Bernoulli numbers}, Rev. R. Acad. Cienc. Exactas Fís. 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 2

In the paper [1] below, the following increasing property for the ratio of two non-zero neighbouring Bernoulli numbers $B_{2n}$ were obtained:

1. The sequence $\frac{|B_{2(n+1)}|}{|B_{2n}|}$ is increasing in $n\in\mathbb{N}_0=\{0,1,2,\dotsc\}$ and tends to $+\infty$ as $n\to+\infty$. Consequently, the sequence $|B_{2n}|$is logarithmically convex in $n\in\mathbb{N}_0$.
2. For fixed $\ell\in\mathbb{N}=\{1,2,3,\dotsc\}$, the sequence \begin{equation}\label{Bernou-ratio-frac-seq} \frac{\prod_{k=1}^{\ell}[2(n+1)+k]}{\prod_{k=1}^{\ell}(2n+k)}\frac{|B_{2(n+1)}|}{|B_{2n}|} \end{equation} is increasing in $n\in\mathbb{N}$ and tends to $+\infty$ as $n\to+\infty$. Consequently, for fixed $\ell\in\mathbb{N}$, the sequence \begin{equation}\label{Bernou-frac-seq-log-conv} \frac{(2n+\ell)!}{(2n)!}|B_{2n}| \end{equation} is logarithmically convex in $n\in\mathbb{N}$.

I think these increasing properties established in the paper [1] below are an answer to this question.

Reference

1. Y. Shuang, B.-N. Guo, and F. Qi, Logarithmic convexity and increasing property of the Bernoulli numbers and their ratios, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM 115 (2021), no. 3, Paper No. 135, 12 pages; available online at https://doi.org/10.1007/s13398-021-01071-x.

Answer 3

In theory, if obtaning a closed-form formula of the Bernoulli polynomials $B_n(x)$, one can compute the ratio $\frac{B_{j+1}(x)}{B_j(x)}$ and its properties, including its asymptotic behaviours, its bounds, and its recursive relations.

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}

I think that, by these closed-form formulas of the Bernoulli polynomials $B_n(x)$, it is not much hard to confirm that the ratio $\frac{B_{j+1}(x)}{B_j(x)}$ cannot be bounded as $j\to+\infty$.

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.