﻿ An Elementary Proof of <img src=image/tit1.png></img> and a Recurrence Formula for ζ(2<i>k</i>)
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### An Elementary Proof of and a Recurrence Formula for ζ(2k)

F. M. S. Lima
Turkish Journal of Analysis and Number Theory. 2017, 5(4), 143-145. DOI: 10.12691/tjant-5-4-5
Received May 27, 2017; Revised June 27, 2017; Accepted June 27, 2017

### Abstract

In this note, a series expansion technique introduced recently by Dancs and He for generating Euler-type formulae for odd zeta values ζ(2k+1), ζ(s) being the Riemann zeta function and k a positive integer, is modified in a manner to furnish the even zeta values ζ(2k). As a result, we find an elementary proof of , as well as a recurrence formula for ζ(2k) from which it follows that the ratio ζ(2k)/π2k is a rational number, without making use of Euler's formula and Bernoulli numbers.

### 1. Introduction

For real values of , , the Riemann zeta function is defined as . In this domain, this series converges according to the integral test.{1} For , , , Euler (1740) did find that 4

 (1)

where is the k-th Bernoulli number, i.e. the rational coefficient of in the Taylor series expansion of , . As a consequence, since one has , which is the Euler solution to the Basel problem (see Ref. 2 and references therein).

In Ref. 3, Dancs and He (2006) introduced a series expansion approach to derive Euler-type formulae for . On noting that their approach could be modified in a manner to furnish similar formulas for , here in this note we show that the change of by in the Dancs-He initial series in fact yields a series expansion which can be reduced to a finite sum involving only even zeta values. From the first few terms of this sum, we have found an elementary proof of and a recurrence formula for . The proofs are elementary in the sense they do not involve complex analysis, Fourier series, or multiple integrals.{2}

### 2. Elementary Evaluation of ζ(2)

For any real and , we begin by taking into account the following Taylor series expansion considered by Dancs and He in Ref. 3:

 (2)

which converges absolutely for .

From the generating function for the Euler polynomial , i.e. , it is clear that , for all nonnegative integer values of m. For , we have

 (3)

Let us take this series as our definition of , being a positive integer. Therefore

 (4)

for all integer .

Now, let

be an auxiliary function, with u belonging to the same domain as above. Since , then can be written in the form

On expanding in a Taylor series, one finds

in which the change of sums justifies by Fubini's theorem. By writing the last series in terms of , one has

 (5)

This is enough for the derivation of our first result.

Theorem 1 (Short evaluation of ζ(2)). .

Proof. By taking the limit as on both sides of Eq.(5), one has

 (6)

which, in face of the value of stated in Eq.(4), implies that

 (7)

Since and for all , the right-hand side of this equation reduces to ,{3} which implies that

and then .

### 3. Recurrence Formula for ζ(2k)

Interestingly, our approach can be easily adapted to treat higher zeta values by changing the exponent of n from 2 to 2k. The result is the following recurrence formula for even zeta values.

Theorem 2 (Recurrence for ζ(2k)). For any positive integer k,

Proof. We begin by defining . Again, since , we may write

 (8)

On rewriting the last series in terms of , one has

Now, on substituting in the above series, one finds

 (9)

The limit as , taken on both sides of Eq.(9), yields

 (10)

From Eq.(4), one knows that

For nonnegative values of m, one has , the only exception being . This reduces Eq.(10) to

By extracting the last term of the sum and isolating ζ(2k), one finds

A multiplication by 2 on both sides yields the desired result.

The first few even zeta values can be readily obtained from the recurrence formula in Theorem 2. For , the sum in the right-hand side is null and our recurrence reduces to

which simplifies to , in agreement to our Theorem 1. For , our recurrence yields

By substituting and multiplying both sides by 4, one finds

 (11)

which implies that .

Note that, by writing the recurrence formula in Theorem 2 in the form

 (12)

it is straightforward to show, by induction on k, that the ratio is a rational number for every positive integer k, without making use of Euler's formula for , as stated in Eq.(1), and Bernoulli numbers. In fact, this was the original motivation that has led the author to study the properties of the Dancs-He series expansions. The proofs developed here could well be modified to cover other special functions of interest in analytic number theory.

### Notes

1. For s=1, one has the harmonic series , which diverges to infinity.

2. For non-elementary proofs, see, e.g., Refs. [1,5] and references therein.

3. This occurs because E2m(1) = 0 for all positive integer values of m.

### References

 [1] M. Aigner and G.M. Ziegler, Proofs from THE BOOK, 5th ed., Springer, New York, 2014, Chap. 9. In article View Article [2] R. Ayoub, “Euler and the zeta function,” Am. Math. Monthly 81, 1067-1085 (1974). In article View Article [3] M. J. Dancs and T.-X. He, “An Euler-type formula for ζ(2k+1),” J. Number Theory 118, 192-199 (2006). In article View Article [4] L. Euler, “De summis serierum reciprocarum,” Commentarii Academiae Scientiarum Petropolitanae 7, 123-134 (1740). In article [5] D. Kalman, “Six ways to sum a series,” Coll. Math. J. 24, 402-421 (1993). In article View Article