Another elementary proof of $: sum_{n ge 1}{1/{n^2}} = pi^2/6,$ and a recurrence formula for $,zeta{(2k)}$

where B k is the k-th Bernoulli number. 2 As a consequence, since B 2 = 1/6 one has ζ(2) = π 2 /6, which is the Euler solution to the Basel problem (see Ref. [2] and references therein). By noting that the series expansion approach introduced by Dancs and He (2006) on seeking for an Euler-type formula for ζ(2k + 1), see Ref. [3], could be modified in a manner to furnish similar formulas for ζ(2k), here in this note Email address: fabio@fis.unb.br (F. M. S. Lima) 1 In this domain, this series converges according to the integral test. For s = 1, one has the harmonic series ∞ n=1 1/n, which diverges to infinity. 2 The (rational) numbers B k are the coefficients of z k /k! in the Taylor series expansion of z/(e z -1), |z| < 2 π.

I show that the substitution of sin (nπ) by cos (nπ) 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, I have found an elementary proof of ζ(2) = π 2 /6 and a recurrence formula for ζ(2k). The proofs are elementary in the sense they do not involve complex analysis, Fourier series, or multiple integrals. 3

For any real ǫ > 0 and u ∈ [1, 1 + ǫ], we begin by taking into account the following Taylor series expansion considered by Dancs and He in Ref. [3]:

which converges absolutely for |t| < π.

From the generating function for the Euler polynomial E m (x), namely 2 e x t /(e t + 1) =

, for all nonnegative integer values of m. For u > 1, we have

Let us take this series as our definition of φ -m (u), m being a positive integer. Therefore

for all integer m > 1. Now, let

be an auxiliary function, with u belonging to the same domain as above. Since cos (nπ) = (-1) n , then f (u) can be written in the form

On expanding cos (nπ) in a Taylor series, one has

in which the change of sums justifies by Fubini’s theorem. By writing the last series in terms of φ m (u), one has

This is sufficient for proofing our first result.

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

Proof. By taking the limit as u → 1 + on both sides of Eq. ( 5), one has lim

which, in face of the value of φ -2 (1) stated in Eq. ( 4), implies that

Since E 0 (1) = 1 and E m (1) = 0 for all m > 0, the right-hand side of this equation reduces to -1 2 ζ(2) + π 2 /4, which implies that

and then

Interestingly, our approach can be easily adapted to treat higher even 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,

which implies that ζ(4) = π 4 /90. Note that, by writing the recurrence formula in Theorem 2 in the form

it is straightforward to show, by induction on k, that the ratio ζ(2k)/π 2k is a rational number for every positive integer k, without making use of Euler’s formula for ζ(2k), see 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.

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