# Where to find or how to prove that the ratio of two Bernoulli polynomials is increasing?

It is well known that the classical Bernoulli polynomials $$B_j(t)$$ are generated by $$\begin{equation*} \frac{s\operatorname{e}^{ts}}{\operatorname{e}^s-1}=\sum_{j=0}^{\infty}B_j(t)\frac{s^j}{j!}, \quad |s|<2\pi. \end{equation*}$$ In my study, I need the conclusion:

The ratio $$\Bigl|\frac{B_{2n+1}(t)}{B_{2n+3}(t)}\Bigr|$$ for $$n\in\{0\}\cup\mathbb{N}$$ is decreasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$ and increasing in $$t\in\bigl(\frac12,1\bigr)$$.

But I do not know where to find this conclusion.

Could you please give a reference to this conclusion? Could you please provide a proof of this conclusion?

• Since $B_{m}(1-t)=(-1)^m B_{m}(t)$ for $m\ge0$, it is sufficient to prove the first part: the ratio $\frac{B_{2m+1}(t)}{B_{2m+3}(t)}$ for $m\in\{0\}\cup\mathbb{N}$ is increasing in $t\in\bigl(0,\frac{1}{2}\bigr)$. Commented Jan 3 at 13:31

$$\newcommand{\s}{\overset{\text{sign}}=}$$Let us first recall some facts about the Bernoulli polynomials $$B_k(x)$$ and the Bernoulli numbers, which latter will be denoted here by $$b_k$$: $$\begin{equation*} B_k(x)=\sum_{j=0}^k\binom nk b_{n-k}x^k \tag{1}\label{1} \end{equation*}$$ -- see e.g. this; $$\begin{equation*} b_0=1,\ b_1=-1/2,\ b_{2n+1}=0; \tag{2}\label{2} \end{equation*}$$ $$\begin{equation*} b_{2n}={\frac {(-1)^{n+1}2(2n)!}{(2\pi )^{2n}}}\zeta(2n) \tag{3}\label{3} \end{equation*}$$ -- see e.g. this, so that $$\begin{equation*} b_{2n}\s(-1)^{n+1}; \tag{4}\label{4} \end{equation*}$$ $$\begin{equation*} B'_k=kB_{k-1} \tag{5}\label{5} \end{equation*}$$ -- see e.g. formulas (1.1) and (1.5) in this paper by Leeming. Everywhere here, by default $$k$$ and $$n$$ are positive integers, unless specified otherwise; and $$\s$$ denotes the equality in sign.

Next -- see e.g. p. 125 in Leeming's paper, $$\begin{equation*} B_{2n+1}(0)=B_{2n+1}(1/2)=0,\quad B_{2n+1}(x)\ne0\text{ for }x\in(0,1/2), \tag{6}\label{6} \end{equation*}$$ $$\begin{equation*} \exists! x_{2n}\in(0,1/2)\ B_{2n}(x_{2n})=0. \tag{7}\label{7} \end{equation*}$$ It follows from \eqref{5}, \eqref{1}, and \eqref{4} that $$B'_{2n+1}=(2n+1)B'(0)\s(-1)^{n+1}$$ and hence, by \eqref{6}, $$\begin{equation*} B_{2n+1}(x)\s(-1)^{n+1}\text{ for }x\in(0,1/2). \tag{8}\label{8} \end{equation*}$$ Next, by \eqref{5} and \eqref{8}, for $$n\ge1$$ we have $$B'_{2n}\s B_{2n-1}\s(-1)^n$$ (on $$(0,1/2)$$). Also, $$B_{2n}B'_{2n}>0$$ on $$(x_{2n},1)$$ and $$B_{2n}B'_{2n}<0$$ on $$(0,x_{2n})$$. So, $$\begin{equation*} B_{2n}\s \begin{cases} (-1)^{n+1}&\text{ on }(0,x_{2n}), \\ (-1)^n&\text{ on }(x_{2n},1). \end{cases} \tag{9}\label{9} \end{equation*}$$ Moreover, it was shown by Ostrowski (see again, e.g., p. 125 in Leeming's paper) that $$x_{2n} for all $$n\ge1$$.

So, letting
$$\begin{equation*} r_k:=\frac{B_k}{B_{k+2}}. \tag{10}\label{10} \end{equation*}$$ we have $$\begin{equation*} r_{2n} \begin{cases} >0&\text{ on }(x_{2n},x_{2n+2})\ne\emptyset, \\ <0&\text{ on }(x_{2n+2},1) \end{cases} \end{equation*}$$ and hence $$\begin{equation*} r_{2n}\to \begin{cases} \infty &\text{ as }x\uparrow x_{2n+2}, \\ -\infty &\text{ as }x\downarrow x_{2n+2}. \end{cases} \tag{10.5}\label{10.5} \end{equation*}$$

After these preliminaries, we see that have to show that $$|r_{2n+1}|$$ is decreasing on $$(0,1/2)$$.
In view of \eqref{8}, $$r_{2n+1}<0$$ (everywhere on $$(0,1/2)$$). So, our task is to show that $$r_{2n+1}$$ is increasing on $$(0,1/2)$$.

We will show a bit more: \begin{equation*} \begin{aligned} &\text{If k is odd, then r_k is increasing on (0,1/2).} \\ &\text{If k is even, then r_k is increasing on (0,x_{k+2}) and on (x_{k+2},1/2).} \end{aligned} \tag{\dagger}\label{dagger} \end{equation*}

Here $$x_{k+2}$$ is defined by (7), as the only zero of $$B_{k+2}$$ in the interval $$(0,1/2)$$.

Note that, in view of \eqref{6}, $$r_k$$ is continuous on $$(0,1/2)$$ if $$k$$ is odd (and, by \eqref{7}, $$r_k$$ has the only point of discontinuity, at $$x_{k+2}$$, if $$k$$ is even). Claim \eqref{dagger} is illustrated below by the graphs $$\{(x,r_k(x))\colon0 for $$k=3$$ (left) and $$k=4$$ (right, with only part of the graph shown, because of the infinite discontinuity at $$x_{k+2}=x_6$$):

The proof of \eqref{dagger} will be done by induction on $$k$$. The induction base, for $$k=0$$ and $$k=1$$, is checked easily, since $$B_0(x)=1$$, $$B_1(x)=x-1/2$$, $$B_2(x)=1/6 - x + x^2$$, and $$B_3(x)=x/2 - 3 x^2/2 + x^3$$.

Suppose now that the statement \eqref{dagger} holds for some integer $$k\ge1$$. We then have to show that \eqref{dagger} holds with $$k+1$$ in place of $$k$$. This will be accomplished using so-called l'Hospital-type rules for monotonicity (l'H). To use these rules, consider the "derivative ratio" for $$r_{k+1}$$ (cf. \eqref{10}): $$\begin{equation*} \rho_{k+1} :=\frac{B'_{k+1}}{B'_{k+3}}=\frac{k+1}{k+3}\frac{B_k}{B_{k+2}}=\frac{k+1}{k+3}\,r_k, \end{equation*}$$ by \eqref{5} and \eqref{10}, so that the monotonicity pattern of $$\rho_{k+1}$$ is the same as that of $$r_k$$: \begin{equation*} \begin{aligned} &\text{If k is odd, then \rho_{k+1} is increasing on (0,1/2).} \\ &\text{If k is even, then \rho_{k+1} is increasing on (0,x_{k+2}) and on (x_{k+2},1/2).} \end{aligned} \tag{11}\label{11} \end{equation*}

Consider first the case when $$k=2n$$, even. Then, by Proposition 4.1 in the l'H paper, \eqref{11}, and the condition $$B_{2n+1}(0)=B_{2n+1}(1/2)=0$$ in \eqref{6}, we see that \eqref{dagger} implies that $$r'_{k+1}>0$$ on $$(0,x_{k+2})$$ and on $$(x_{k+2},1/2)$$, so that the induction step is done in this case.

Consider now the case when $$k=2n-1\ge1$$, odd. Then it is not true that $$B_{k+1}(0)=0$$ or $$B_{k+1}(1/2)=0$$. So, in this case, we have to use so-called general l'Hospital-type rules for monotonicity. More specifically, we will use Table 1.1, with $$f=B_{k+1}=B_{2n}$$ and $$g=B_{k+3}=B_{2n+2}$$, so that $$r_{k+1}=f/g$$ and, by \eqref{5} and \eqref{8}, $$g'\s B_{2n+1}\s(-1)^{n+1}$$ on the entire interval $$(0,1/2)$$. So, by \eqref{7}, $$gg'\s B_{2n+2}B_{2n+1}\s B_{2n+2}(-1)^{n+1}$$, which, by \eqref{9}, is $$<0$$ on the interval $$(0,x_{2n+2})=(0,x_{k+3})$$ and $$>0$$ on the interval $$(x_{2n+2},1/2)=(x_{k+3},1/2)$$. Hence, by
lines 3 and 1 of mentioned Table 1.1 and \eqref{11}, we get that $$r_{k+1}=r_{2n}$$ is

• up-down on $$(0,x_{k+3})$$ (that is, for some $$c_1\in[0,x_{k+3}]$$ the function $$r_{k+1}=r_{2n}$$ is increasing on $$(0,c_1)$$ and decreasing on $$(c_1,x_{k+3})$$;

• down-up on $$(x_{k+3},1/2)$$ (that is, for some $$c_2\in[x_{k+3},1/2]$$ the function $$r_{k+1}=r_{2n}$$ is decreasing on $$(x_{k+3},c_2)$$ and increasing on $$(c_2,1/2)$$.

However, in view of \eqref{10.5}, $$r_{2n}$$ cannot be decreasing in any left neighborhood of $$x_{k+3}=x_{2n+2}$$, and $$r_{2n}$$ cannot be decreasing in any right neighborhood of $$x_{k+3}=x_{2n+2}$$. So, in the above "up-down" and "down-up" items, $$c_1=x_{k+3}=c_2$$; that is, $$r_{k+1}=r_{2n}$$ is increasing on $$(0,x_{k+3})$$ and on $$(x_{k+3},1/2)$$.

Thus, whether $$k$$ is even or odd, \eqref{dagger} holds with $$k+1$$ in place of $$k$$, which completes the induction step. $$\quad\Box$$

• Thanks for this answer! I believe that there should be other short answers to this problem. Commented Jan 5 at 4:27
• @qifeng618 : Is there any particular reason to believe that? Commented Jan 5 at 4:28
• By my own instinct. Commented Jan 5 at 4:32
• @qifeng618 : I would be much surprised if there is a significantly shorter solution. Indeed, in view of (5), it seems natural to consider the derivative ratio, as was done just before (11). The complications that come then are also natural, because the even-numbered Bernoulli polynomials $B_{2n}$ do not vanish at the endpoints but vanish inside the interval; plus, the Bernoulli polynomials are nontrivial objects. Commented Jan 5 at 5:19
• @qifeng618 : It's been over a week since this answer was posted. There has been no other activity on this page, not even a single comment. Moreover, I explained why the presented solution is quite natural. It certainly seems unlikely at this point that any other answer is forthcoming. So, please be fair and finalize this matter. If something transpires later, you can always reconsider. Commented Jan 12 at 20:09

This answer is a slight revision of the proof of Proposition 1 in Section 4 of the forthcoming paper:

Gui-Zhi Zhang, Zhen-Hang Yang, and Feng Qi, On normalized tails of series expansion of generating function of Bernoulli numbers, Proceedings of the American Mathematical Society (2024), in press; available online at https://www.researchgate.net/publication/379248377 or at https://doi.org/10.1090/proc/16877.

Lemma. For $$a,b\in\mathbb{R}$$ with $$a, let $$p(s)$$ and $$q(s)$$ be continuous on $$[a,b]$$, differentiable on $$(a,b)$$, and $$q'(s)\ne 0$$ on $$(a,b)$$. If the ratio $$\frac{p'(s)}{q'(s)}$$ is increasing in $$s\in(a,b)$$, then both $$\frac{p(s)-p(a)}{q(s)-q(a)}$$ and $$\frac{p(s)-p(b)}{q(s)-q(b)}$$ are increasing in $$s\in(a,b)$$.

Theorem. The ratio $$\frac{B_{2m-1}(t)}{B_{2m+1}(t)}$$ for $$m\in\mathbb{N}$$ is increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$ and decreasing in $$t\in\bigl(\frac{1}{2},1\bigr)$$.

Proof. The inequality $$\begin{equation*} (-1)^{m+1}B_{2m+1}(t)>0, \quad m\ge0, \quad 0 means that the ratio $$\frac{B_{2m-1}(t)}{B_{2m+1}(t)}$$ for $$m\in\mathbb{N}$$ is negative in $$t\in\bigl(0,\frac{1}{2}\bigr)$$. The identity $$\begin{equation*} B_{m}(1-t)=(-1)^m B_{m}(t), \quad m\ge0 \end{equation*}$$ means that $$B_{2m-1}\bigl(\frac{1}{2}\bigr)=0$$ for $$m\in\mathbb{N}$$ and the ratio $$\frac{B_{2m-1}(t)}{B_{2m+1}(t)}$$ for $$m\in\mathbb{N}$$ is symmetric with respect to $$t=\frac{1}{2}$$ on $$(0,1)$$. Hence, it is sufficient to prove the increasing property only on $$\bigl(0,\frac{1}{2}\bigr)$$.

It is well known that $$B_m(0)=B_m$$ for $$m\ge0$$ and $$$$\label{B(n)-explit-form}\tag{Y0} B_m(t)=\sum_{k=0}^{m}\binom{m}{k}B_k t^{m-k}, \quad m\ge0.$$$$ This means that the function $$\begin{equation*} f_{2m-1}(t)=\frac{B_{2m-1}(t)}{t}=\sum_{k=0}^{2m-1}\binom{2m-1}{k}B_k t^{2m-k-2}\ne0 \end{equation*}$$ in $$t\in\bigl(0,\frac{1}{2}\bigr)$$ and $$f_{2m-1}\bigl(\frac{1}{2}\bigr)=0$$ for $$m\in\mathbb{N}$$.

Directly differentiating leads to $$\begin{equation*} f_{2m-1}'(t)=\frac{t B_{2m-1}'(t)-B_{2m-1}(t)}{t^2} =\frac{(2m-1)t B_{2m-2}(t)-B_{2m-1}(t)}{t^2} \end{equation*}$$ for $$m\in\mathbb{N}$$ and $$$$\label{deriv-deriv-eq}\tag{Y1} h_{2m-2}'(t)=\bigl[t^2 f_{2m-1}'(t)\bigr]'=(2m-1)(2m-2)t B_{2m-3}(t)\ne0, \quad m\ge2$$$$ in $$t\in\bigl(0,\frac{1}{2}\bigr)$$, where we used the formula $$\begin{equation*}%\label{Bernou-polyn-deriv} B_m'(t)=m B_{m-1}(t), \quad m\in\mathbb{N}. \end{equation*}$$ Since $$\begin{equation*} \lim_{t\to0}\bigl[t^2 f_{2m-1}'(t)\bigr]=0,\quad m\ge2, \end{equation*}$$ the equation \eqref{deriv-deriv-eq} implies that $$f_{2m-1}'(t)\ne0$$ for $$m\in\mathbb{N}$$ in $$t\in\bigl(0,\frac{1}{2}\bigr)$$.

It is easy to see that $$\begin{gather} \label{Bern-deriv-ratio-eq}\tag{Y2} \frac{B_{2m-1}(t)}{B_{2m+1}(t)}=\frac{f_{2m-1}(t)}{f_{2m+1}(t)} =\frac{f_{2m-1}(t)-f_{2m-1}\bigl(\frac{1}{2}\bigr)}{f_{2m+1}(t)-f_{2m+1}\bigl(\frac{1}{2}\bigr)},\\ \frac{f_{2m-1}'(t)}{f_{2m+1}'(t)}=\frac{(2m-1)t B_{2m-2}(t)-B_{2m-1}(t)} {(2m+1)t B_{2m}(t)-B_{2m+1}(t)} =\frac{h_{2m-2}(t)-h_{2m}(0)}{h_{2m}(t)-h_{2m}(0)}, \label{f-deriv-ratio-eq}\tag{Y3} \end{gather}$$ and $$$$\label{h-deriv-ratio}\tag{Y4} \frac{h_{2m-2}'(t)}{h_{2m}'(t)}=\frac{(2m-1)(2m-2)} {(2m+1)(2m)}\frac{B_{2m-3}(t)}{B_{2m-1}(t)}$$$$ for $$m\ge2$$, where $$\begin{equation*} h_{2m-2}(t)=t^2 f_{2m-1}'(t)=(2m-1)t B_{2m-2}(t)-B_{2m-1}(t)\ne0 \end{equation*}$$ for $$t\in\bigl(0,\frac{1}{2}\bigr)$$ and $$h_{2m-2}(0)=0$$ for $$m\ge2$$.

From the formula \eqref{B(n)-explit-form}, it follows that $$\begin{equation*} \frac{B_1(t)}{B_3(t)}=\frac{1}{(t-1) t} \quad\text{and}\quad \frac{B_3(t)}{B_5(t)}=\frac{3}{3 t^2-3 t-1}. \end{equation*}$$ Directly differentiating yields $$\begin{equation*} \biggl[\frac{B_1(t)}{B_3(t)}\biggr]'=\frac{1-2 t}{(t-1)^2 t^2} \quad\text{and}\quad \biggl[\frac{B_3(t)}{B_5(t)}\biggr]'=\frac{9 (1-2 t)}{(3 t^2-3 t-1)^2}. \end{equation*}$$ Accordingly, the ratios $$\frac{B_1(t)}{B_3(t)}$$ and $$\frac{B_3(t)}{B_5(t)}$$ are both increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$.

Assume that the ratio $$\frac{B_{2m-1}(t)}{B_{2m+1}(t)}$$ for some $$m\ge3$$ is increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$. Then, by \eqref{h-deriv-ratio}, we see that the derivative ratio $$\begin{equation*} \frac{h_{2m}'(t)}{h_{2m+2}'(t)}=\frac{(2m+1)m} {(2m+3)(m+1)}\frac{B_{2m-1}(t)}{B_{2m+1}(t)}, \quad m\ge3 \end{equation*}$$ is also increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$. Applying the monotonicity rule stated in the above lemma on the interval $$\bigl(0,\frac{1}{2}\bigr)$$ and utilizing \eqref{f-deriv-ratio-eq} reveal that the derivative ratio $$\frac{f_{2m+1}'(t)}{f_{2m+3}'(t)}$$ for $$m\ge3$$ is increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$. Applying the monotonicity rule recited in the above lemma on the interval $$\bigl(0,\frac{1}{2}\bigr)$$ once again and employing \eqref{Bern-deriv-ratio-eq} yield that the ratio $$\frac{B_{2m+1}(t)}{B_{2m+3}(t)}$$ for $$m\ge3$$ is increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$.

By mathematical induction, we proved that the ratio $$\frac{B_{2m-1}(t)}{B_{2m+1}(t)}$$ for $$m\in\mathbb{N}$$ is increasing in $$t\in\bigl(0,\frac{1}{2}\bigr)$$. The required proof is complete.

References

1. M. Abramowitz and I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, Reprint of the 1972 edition, Dover Publications, Inc., New York, 1992.
2. G. D. Anderson, M. K. Vamanamurthy, and M. Vuorinen, Conformal Invariants, Inequalities, and Quasiconformal Maps; John Wiley & Sons: New York, NY, USA, 1997.
3. F. W. J. Olver, D. W. Lozier, R. F. Boisvert, and C. W. Clark (eds.), NIST Handbook of Mathematical Functions, Cambridge University Press, New York, 2010; available online at http://dlmf.nist.gov/.
• Except this proof and Iosif Pinelis’s proof, there has been the third proof which was given by a colleague of mine. Commented Mar 25 at 10:47
• In a forthcoming paper, Yang and Qi establish the monotonicity of the ratios \begin{equation*} \frac{B_{2n-1}(t)}{B_{2n+1}(t)}, \quad \frac{B_{2n}(t)}{B_{2n+1}(t)},\quad \frac{B_{2m}(t)}{B_{2n}(t)},\quad \frac{B_{2n}(t)}{B_{2n-1}(t)} \end{equation*} in $t\in\bigl(0,\frac12\bigr)$ and derive some new and known inequalities of the Bernoulli polynomials $B_n(t)$, the Bernoulli numbers $B_{2n}$, and their ratios such as $\frac{B_{2n+2}}{B_{2n}}$. Commented May 4 at 2:03
• About more new results and applications of the ratios of two Bernoulli polynomials, please refer to the arXiv preprint at arxiv.org/abs/2405.05280. In my opinion, this is an interesting problem. Commented May 10 at 0:47