The Mathematics Student ISSN: 0025-5742 Vol. 88, Nos. 1-2, January-June (2019), xx{yy A SIMPLE PROOF THAT ζ(2) = π2=6 M. RAM MURTY (Received: 22 - 03 - 2019) Abstract. We give a simple proof that ζ(2) = π2=6 using only first- year calculus. An elementary recursion allows us to deduce that gen- 2k erally, ζ(2k) 2 π Q, a celebrated theorem of Euler. 1. Introduction In 1650, Pietro Mengoli posed the problem of explicitly evaluating 1 1 1 1 + + + + ··· 22 32 42 and this had come to be known as the Basel problem. The solution is π2=6 and it was discovered by Euler almost a century later in 1734, when Euler was 28 years old. Euler's proof, though correct, was far from rigorous and had to await further developments in complex analysis to put it on a sure footing. We will give a slick proof that uses only ideas from first year 2k calculus. In an equally simple way, we will show that ζ(2k) 2 π Q, where ζ(s) is the Riemann zeta function which for s > 1 is given by X1 1 ζ(s) = : n=1 ns 2. The Proof We first observe that Z 1 2y 2y Z N 2y 2y 2 − 2 2 dy = lim 2 − 2 2 dy 0 1 + y x + y N!1 0 1 + y x + y h 1 + y2 iy=1 = log = 2 log x; (2.1) x2 + y2 y=0 since, for fixed x, 1 + y2 lim log = 0: y!1 x2 + y2 Simplifying the integrand in (2.1), we deduce log x Z 1 ydy − 2 = 2 2 2 : (2.2) 1 − x 0 (1 + y )(x + y ) 2010 Mathematics Subject Classification: 11M06, 11M32 Key words and phrases: Riemann zeta function, zeta values. c Indian Mathematical Society, 2019 . 1 2 M. RAM MURTY Now consider the integral Z 1 log x I = 2 dx: 0 1 − x Expanding 1=(1 − x2) as a power series and integrating by parts term by term, we get Z 1 X1 2n X1 1 I = x (log x)dx = − 2 : n=0 0 n=0 (2n + 1) Thus 1 1 1 I = − 1 + + + ··· = − 1 − ζ(2): (2.3) 32 52 22 Changing x to 1=x in the integral leads to Z 1 log 1=x dx Z 1 log x I = 2 − 2 = 2 dx 1 1 − (1=x ) x 1 1 − x so that Z 1 log x 2I = 2 dx: (2.4) 0 1 − x Inserting (2.2) into (2.4), we deduce that Z 1 Z 1 ydxdy −2I = 2 2 2 : 0 0 (1 + y )(x + y ) In the inner integral, changing x to yx leads to Z 1 Z 1 dxdy 2 − 2I = 2 2 = π =4 (2.5) 0 0 (1 + x )(1 + y ) since Z 1 dx 1 2 = [arctan x]0 = π=2: 0 1 + x Combining (2.5) with (2.3) gives the desired result. 3. The formula for ζ(2k) 2k It is possible to show ζ(2k) 2 π Q using the following simple recursion which certainly goes back to Euler. Theorem 1. For k ≥ 2, Xk−1 (k + (1=2))ζ(2k) = ζ(2i)ζ(2k − 2i): (3.1) i=1 2k This gives a recursion from which ζ(2k) 2 π Q is easily deduced. Our proof of this relies on the following lemma. Lemma 2. For any natural number n, X01 1 = 3=4n2; m=1 m2 − n2 where the dash on the summation means that we exclude m = n. A SIMPLE PROOF THAT ζ(2) = π2=6 3 Proof. The sum is 1 X0N 1 1 lim − N!1 2n m=1 m − n m + n 1 Xn−1 1 XN−n 1 XN+n 1 1 = lim − + − + : N!1 2n j=1 j j=1 j j=n+1 j 2n Using the elementary fact that as N ! 1, XN 1 = log N + γ + o(1); j=1 j where γ is Euler's constant, the sum in the brackets is N − n 3 log + + o(1) N + n 2n 2 so that our limit is 3=4n , as claimed. We can now prove Theorem 1. Inserting the series for the zeta function, the right hand side of (3.1) is X1 X1 Xk−1 1 1 : n=1 m=1 i=1 n2i m2k−2i When n 6= m, the innermost sum being a geometric sum, is easily seen to be (n2−2k − m2−2k)=(m2 − n2): Separating the contribution from m = n, we deduce that our sum is X1 X01 1 1 1 (k − 1)ζ(2k) + − : n=1 m=1 n2k−2 m2k−2 m2 − n2 Using the lemma, we see that the double sum contributes (3=2)ζ(2k), which gives the result as claimed. M. Ram Murty Dept. of Mathematics and Statistics Queen's University, Kingston, Ontario, Canada, K7L 3N6 E-mail: [email protected] 4.