Bernoulli and Euler numbers from divergent series

TL;DR

This paper offers a straightforward proof of identities involving Bernoulli and Euler numbers using Abel sums of divergent series, emphasizing the role of linearity in summation methods.

Contribution

It introduces a simple proof technique for classical identities of Bernoulli and Euler numbers via Abel summation of divergent series.

Findings

Identifies identities relating Bernoulli and Euler numbers.

Demonstrates the use of Abel sums for divergent series.

Highlights the importance of linearity in summation methods.

Abstract

The aim of this note is to provide a simple proof of some well-known identities and recurrences relating classical Bernoulli and Euler numbers by using the Abel sum of the divergent series $\sum_{n=0}^\infty (-1)^{n} (n+1)^k$, $k$ a positive integers. Special attention is placed on the fact that the numerical value of these sums is determined by the linearity of the summation method involved.

Equations

$$\frac{z}{e^z-1}=-\frac{z}{2}+\frac{z}{2}\frac{e^z+1}{e^z-1}=\sum _{k=0}^{\infty }\frac{B_k}{k!}{z}^k,|z|<2\pi .$$

$$\sec (z)=\frac{1}{\cos (z)}=\sum _{n=0}^{\infty }\frac{E_n}{n!}{z}^n=1+\sum _{n=1}^{\infty }\frac{E_{2n}}{(2n)!}{z}^{2n},|z|<\frac{\pi }{2}.$$

$$z\cot (z)=\sum _{k=0}^{\infty }\frac{B_{2k}}{(2k)!}(-1{)}^k{2}^{2k}{z}^{2k},|z|<\pi ,$$

$$\tan (z)=\sum _{n=1}^{\infty }(-1{)}^{n+1}\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}{z}^{2n-1}=-2i\sum _{k=1}^{\infty }\frac{2^{k+1}-1}{k+1}{B}_{k+1}\frac{(2iz{)}^k}{k!},|z|<\frac{\pi }{2}.$$

$$-\frac{1}{1+e^z}=\sum _{k=0}^{\infty }\frac{2^{k+1}-1}{k+1}{B}_{k+1}\frac{z^k}{k!},|z|<\pi .$$

$$\frac{2^{k+1}-1}{k+1}{B}_{k+1}=\frac{1}{2}-\sum _{l=1}^k\left(\genfrac{}{}{0ex}{}{k}{l}\right)\frac{2^{l+1}-1}{l+1}{B}_{l+1},k\ge 1.$$

$$B_{n+1}=\frac{(-1{)}^n(n+1)}{2^{n+1}-1}\sum _{k=1}^{n+1}\frac{1}{2^k}\sum _{j=0}^{k-1}\left(\genfrac{}{}{0ex}{}{k-1}{j}\right)(-1{)}^j(j+1{)}^n,$$

$${\sigma }_k:=1^k-2^k+3^k-4^k+\cdots ,k\in \mathbb{N},$$

$$1^{k}x-2^k{x}^2+3^k{x}^3-\cdots =\sum _{n=0}^{\infty }(-1{)}^n(n+1{)}^k{e}^{-(n+1)y}=(-1{)}^k\frac{d^k}{d{y}^k}\left(\frac{1}{1+e^y}\right),$$

$$A(1^k-3^k+5^k-7^k+\cdots )=(-1{)}^{\lfloor k/2\rfloor }{E}_k/2,$$

$$(1-2+3-4+5-6+\cdots )+(0+1-2+3-4+5-\cdots )=1-1+1-1+1-1+\cdots .$$

$$S^{\ast }({\sigma }_k)=S^{\ast }\left(1-\sum _{n=2}^{\infty }(-1{)}^n(1+(n-1){)}^k\right)=\frac{1}{2}-\sum _{j=1}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)S^{\ast }({\sigma }_j).$$

$$\frac{2^{k+1}-1}{k+1}{B}_{k+1}=\sum _{n=0}^{a-1}(-1{)}^n(n+1{)}^k+\frac{(-1{)}^a}{2}{a}^k+(-1{)}^a\sum _{j=1}^k(\genfrac{}{}{0ex}{}{k}{j})a^{k-j}\frac{2^{j+1}-1}{j+1}{B}_{j+1}.$$

$$A\left(a^{-k}{\sigma }_k\right)$$

$$=\frac{1}{a^k}\sum _{n=0}^{a-1}(-1{)}^n(n+1{)}^k+(-1{)}^{a}A\left(\sum _{n=0}^{\infty }(-1{)}^n\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})\frac{(n+1{)}^j}{a^j}\right).$$

$$\sum _{j=1}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)2^j\frac{2^{j+1}-1}{j+1}{B}_{j+1}=\frac{1}{2}-\frac{(-1{)}^{\lfloor k/2\rfloor }{E}_k}{2},$$

$$2^{k+1}\frac{2^{k+1}-1}{k+1}{B}_{k+1}=\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)(-1{)}^{\lfloor j/2\rfloor }{E}_j,$$

$$A\left(\sum _{n=0}^{\infty }(-1{)}^n(2n+1{)}^k\right)=\frac{1}{2}-\sum _{j=1}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)2^j\frac{2^{j+1}-1}{j+1}{B}_{j+1}.$$

$$A(2^k{\sigma }_k)=A\left(\sum _{n=1}^{\infty }(-1{)}^{n+1}(2n-1+1{)}^k\right)=\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)A\left(\sum _{n=1}^{\infty }(-1{)}^{n+1}(2n-1{)}^j\right).$$

$$\frac{q^k}{2}-\sum _{j=1}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)q^{k-j}{a}^j\frac{2^{j+1}-1}{j+1}{B}_{j+1}=\frac{(-1{)}^k}{2}\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right){\left(\frac{a}{2}-q\right)}^{k-j}{\left(\frac{a}{2}\right)}^j(-1{)}^{\lfloor j/2\rfloor }{E}_j.$$

$$S^{\ast }(\sigma )-S^{\ast }(\sigma )=S^{\ast }(1+2+3+\cdots )-S^{\ast }(0+1+2+\cdots )=S^{\ast }(1+1+\cdots ),$$

$$\begin{array}[]{rllllllllll}\sigma&=&1&+&2&+&3&+&4&+&\cdots,\\ 4\sigma&=&&&4&+&&&8&+&\cdots.\end{array}$$

Full text

The author was supported by the Austrian FWF-Project P 26735-N25.

Many special numeric sequences have been studied intensively due to their appearance and applications in Combinatorics, Number Theory, and Analysis. For instance, Fibonacci, Bernoulli, Euler, Eulerian [1], or Stirling numbers [2]. In addition to common techniques to obtain properties and relations among them, summation of divergent series can be used too!

The Bernoulli and Euler numbers are two sequences of rational numbers that play an important role in mathematical analysis. They appear naturally as the coefficients of the Taylor expansions of trigonometric functions and in the computation of sums of series and asymptotic expansions. For instance, in the calculation of $\zeta(2k)$, where $k$ is an integer and $\zeta$ denotes the Riemann zeta function [3], or in the Euler-Maclaurin summation formula [4] Chapter XIII. They also exhibit interesting relations between other numbers, for example with Euler’s constant [5]. Bernoulli numbers have been called an “unifying force” in mathematics due to their presence in several branches as in Analytic Number Theory [6] and Differential Topology [7].

The Bernoulli numbers $B_{n}$ (with signs) were introduced by J. Bernoulli in his Ars Conjectandi [8, 9], published posthumously in 1713, to give a precise formula for $1^{k}+2^{k}+\cdots+n^{k}$ as a polynomial in $n$. They are defined as the coefficients in the Taylor expansion at $z=0$ of

[TABLE]

From the definition it is deduced that $B_{0}=1$, $B_{1}=-\frac{1}{2}$, $B_{2}=\frac{1}{6}$ and $\sum_{k=0}^{n-1}{n\choose k}B_{k}=0,$ $n\geq 2$. This formula allows to find $B_{n}$ recursively and it shows that $B_{n}$ is always rational. Also, since $\frac{z}{2}\frac{e^{z}+1}{e^{z}-1}$ is an even function, we deduce that $B_{2k+1}=0$, for $k\geq 1$.

The Euler numbers are defined as the Taylor expansion at $z=0$ of

[TABLE]

Then it is clear that $\sum_{k=0}^{n}{2n\choose 2k}(-1)^{k}E_{2k}=0,$ and thus the $E_{2k}$ are integers.

Replacing $z$ by $2iz$ in equation (1) we find the Taylor expansion at $z=0$ of

[TABLE]

and from the identity $\tan(z)=\cot(z)-2\cot(2z)$, we see that

[TABLE]

Thus the Maclaurin series of $\tan(z)$ and $\sec(z)$ are not trivial but there are elementary recursive methods to obtain them [10]. Finally, we obtain from the last equation the power series expansion

[TABLE]

There are many lists of recurrences satisfied by the Bernoulli and Euler numbers, e.g., Nielsen’s classical book [11]. We only need one, namely

[TABLE]

It can be obtained from the equality $\frac{-1}{1+e^{z}}e^{z}=-1+\frac{1}{1+e^{z}}$ and the formula (3) by equating the corresponding coefficients of $z^{k}$.

We will show how to obtain (4) also by summing divergent series. The use of this method is not new. Garabedian [12], for instance, showed that

[TABLE]

from summing

[TABLE]

using Cesàro and Abel summability. Ra̧dkowsk [13] also provided a proof using calculus of finite differences. Similarly, Namias [14] deduced some other recurrences using Stirling’s asymptotic series and the duplication formula for the Gamma function, although the results can be also obtained in an elementary way [15]. We will use the sum of $\sigma_{k}$ and the linearity of a summation method that can sum it to obtain (4) and some other simple recurrences for Bernoulli and Euler numbers.

Let us recall that a series $\sigma=\sum_{n=0}^{\infty}a_{n}$ is said to be Abel summable with sum $A(\sigma)$ if for all $x\in\mathbb{R}$ with $0\leq x<1$, the associated generating series $\sum_{n=0}^{\infty}a_{n}x^{n}$ is convergent and $A(\sigma):=\lim_{x\rightarrow 1^{-}}\sum_{n=0}^{\infty}a_{n}x^{n}$ exists. In particular, the series $\sigma_{k}$ is Abel summable [16, 17]: if we replace $x=e^{-y}$ in the series

[TABLE]

we can use the expansion (3) and take $y\rightarrow 0^{+}$ to find the value $A(\sigma_{k})=\frac{2^{k+1}-1}{k+1}B_{k+1}$, $k\geq 1$. It is also clear that $A(\sigma_{0})=\frac{1}{2}$. In the same way

[TABLE]

where $\lfloor,\rfloor$ denotes the floor function. Indeed, setting $x=e^{-2y}$ in the generating series $\sum_{n=0}^{\infty}(-1)^{n}(2n+1)^{k}x^{n}=(-1)^{k}e^{y}\frac{d^{k}}{dy^{k}}\left(\frac{1}{e^{y}+e^{-y}}\right)$, we can let $y\rightarrow 0^{+}$ and then the formula follows from equation (2).

Is it possible to attribute a sum to a divergent series in a way compatible with the usual rules of calculus? For Euler the answer was positive! This is evidenced in his work De seriebus divergentibus [18] on the Wallis series $\sum_{n=0}^{\infty}(-1)^{n}n!$, where he found the sum $\int_{0}^{\infty}\frac{e^{-t}}{1+t}dt\approx 0.5963473625$ [16]. In the same spirit, this was the belief of Hardy as he exhibited in his book Divergent Series [4]. Nowadays, the theory of summability attempts to answer this question.

We denote by $\mathcal{D}$ the $\mathbb{C}-$vector space of complex sequences $(a_{n})_{n\geq 0}$ and by $\mathcal{C}$ subspace of sequences such that $\lim_{n\rightarrow+\infty}a_{0}+\cdots+a_{n}$ exists. We can think of elements of $\mathcal{D}$ as formal numerical series $\sigma=\sum_{n=0}^{\infty}a_{n}$. The space $\mathcal{C}$ is the domain of the sum homomorphism $S:\mathcal{C}\rightarrow\mathbb{C}$, which associates a series $\sigma$ to its sum $S(\sigma)$, i.e., the limit of its partial sums. From this point of view, a summability method is a map $S^{*}:\mathcal{C}^{\prime}\rightarrow\mathbb{C}$ on some linear subspace $\mathcal{C}\subseteq\mathcal{C}^{\prime}\subseteq\mathcal{D}$ such that the following rules are satisfied:

Regularity rule: If $\sigma\in\mathcal{C}$, then $S^{*}(\sigma)=S(\sigma)$. 2. 2.

Translation rule: $S^{*}\left(\sum_{n=0}^{\infty}a_{n}\right)=a_{0}+S^{*}\left(\sum_{n=1}^{\infty}a_{n}\right)$. 3. 3.

Linearity rule: $S^{*}$ is a $\mathbb{C}-$linear map.

Cesàro and Abel summability are examples of summability methods satisfying such rules. It is worth noting that such axioms were implicitly used by Euler.

We will use the following fact:

Assume that $S^{\ast}$ satisfies the above rules. Then if it sums a series, the value we find through the rules is the value $S^{\ast}$ assigns to the series.

As a first example, we consider $F=1+1+2+3+5+\cdots=\sum_{n=0}^{\infty}F_{n}$, the series of Fibonacci numbers: if $S^{*}$ sums $F$, then $S^{*}({F})=2+S^{*}\left(\sum_{n=2}^{\infty}F_{n}\right)=2+S^{*}\left(\sum_{n=2}^{\infty}F_{n-1}+F_{n-2}\right)=2+(S^{*}({F})-1)+S^{*}({F})$ and thus $S^{*}({F})=-1$. As second example, we take the geometric series $s_{z}=1+z+z^{2}+z^{3}+\cdots$. We have $S^{*}(s_{z})=\frac{1}{1-z}$, $z\neq 1$, since $S^{*}(s_{z})=1+zS^{*}(s_{z})$. In particular, for $z=-1$ we recover the usual value $S^{*}(\sigma_{0})=S^{*}(1-1+1-1+\cdots)=\frac{1}{2}$. The same conclusion is true for any positive integer $k$. This is remarked by Knopp [17, p. 479], but it is not proved there. We include a simple proof using induction on $k$.

[TABLE]

Then by rules 2 and 3, $S^{*}(\sigma_{0})=S^{*}(\sigma_{1})+S^{*}(\sigma_{1})=2S^{*}(\sigma_{1})$ and $S^{*}(\sigma_{1})=\frac{1}{2}S^{*}(\sigma_{0})=\frac{1}{4}$. Now we assume the formula holds for $S^{*}(\sigma_{1}),...,S^{*}(\sigma_{k-1})$, $k\geq 2$. By the binomial theorem, we see that

[TABLE]

Using the induction hypothesis and equation (4), we conclude that the formula is valid for $k$. The principle of induction allows us to conclude the proof. ∎

The previous reasoning provides another way to prove recursion (4): Take $S^{*}=A$ as Abel summability. Since $A$ satisfies rules 2 and 3, we can replace the value $A(\sigma_{k})=\frac{2^{k+1}-1}{k+1}B_{k+1}$ in the recursion obtained in the previous proof. In fact, we can easily generalize equation (4) using the same type of argument.

[TABLE]

[TABLE]

∎

Formula (4) corresponds to the case $a=1$ of Proposition 2. The formula above is simple in the sense that it can be deduced directly from (3) by equating corresponding coefficients of $z^{k}$ in the identity $\sum_{n=0}^{a-1}(-1)^{n}e^{(n+1)z}+(-1)^{a}e^{az}-(-1)^{a}\frac{e^{az}}{e^{z}+1}=1-\frac{1}{1+e^{z}}$.

We can go further and recover the usual formulas to determine the Bernoulli numbers in terms of Euler numbers and vice versa.

[TABLE]

[TABLE]

valid for all integers $k\geq 1$.

[TABLE]

Equating both results we get (6). Similarly, for (7) it is enough to consider the series $2^{k}\sigma_{k}=2^{k}-4^{k}+6^{k}-8^{k}+\cdots$ and the relation

[TABLE]

∎

The formulas (6) and (7) are of course elementary. They can be deduced by equating the coefficients of $z^{k}$ in $e^{z}\frac{-1}{1+e^{2z}}=\frac{-1}{e^{z}+e^{-z}}$, $e^{z}\frac{1}{e^{z}+e^{-z}}=\frac{1}{1+e^{-2z}}$, respectively.

We invite the reader to calculate the Abel sum of $\sum_{n=0}^{\infty}(-1)^{n}(an+q)^{k}$, $k\geq 1$, $a,q\in\mathbb{R}$ and $a>0$ as we did here (setting $x=e^{-ay}$ in the generating series and using rules 2 and 3) to conclude that Bernoulli and Euler numbers also satisfy

[TABLE]

Unfortunately, this expression does not provide new information since it can be deduced directly from (7) by equating the corresponding coefficient of $q^{k-j}a^{j}$.

[TABLE]

and $1+1+1+\cdots$ would be summable with sum equals to [math]. However, there are methods that assign the controversial value $-\frac{1}{12}$ to $\sigma$. For instance, interpreting $\sigma$ as the value of the analytic continuation of $\zeta(z)$ at $z=-1$. Another example is the constant of a series method of Ramanujan [19, 20, 4, p. 327, 346]. Naturally, such methods can not satisfy the rules 1 to 3. It is curious that Ramanujan wrote [19, p. 135]

[TABLE]

Subtracting both equations he found $-3\sigma=1-2+3-4+5-6+\cdots=\frac{1}{4},$ so again $\sigma=-\frac{1}{12}$, although this reasoning is not compatible with our approach.