A simple computation of $ζ(2k)$ by using Bernoulli polynomials and a telescoping series

TL;DR

This work presents an elementary, calculus-based proof of Euler's formula for even zeta-values $\zeta(2k)$ by exploiting the defining properties of Bernoulli polynomials and a telescoping integral trick. It introduces auxiliary integrals $I(k,m)$, derives recurrences via integration by parts, and uses a trigonometric telescope to convert a cosine series into a limit that yields $\zeta(2k)=\frac{(-1)^{k-1}2^{2k-1}\pi^{2k}}{(2k)!} B_{2k}$. The same framework yields an integral representation for odd zeta-values $\zeta(2k+1)$, namely $\zeta(2k+1)=\frac{(-1)^{k-1}2^{2k}\pi^{2k+1}}{(2k+1)!}\int_0^1 B_{2k+1}(t)\cot\left(\frac{\pi t}{2}\right)\,dt$, and discusses connections to harmonic numbers and Ramanujan-type formulas. The paper also clarifies the Fourier-analytic perspective on Bernoulli polynomials and highlights a broadly applicable telescope-based approach that yields a transparent path to zeta-values while illustrating the role of Bernoulli polynomials in these classic identities.

Abstract

We present a new proof of Euler's formulas for $ζ(2k)$, where $k = 1,2,3,...$, which uses only the defining properties of the Bernoulli polynomials, obtaining the value of $ζ(2k)$ by summing a telescoping series. Only basic techniques from Calculus are needed to carry out the computation. The method also applies to $ζ(2k+1)$ and the harmonic numbers, yielding integral formulas for these.

Theorems & Definitions (1)

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