Basel Problem Proof
Cody Dianopoulos
Joran Layne
Alex Yokokawa
June 6, 2013
Abstract
In 1644, Pietro Mengoli questioned the mathematical society about the
sum of the reciprocals of the perfect square numbers. Thanks to Leonhard
Euler in 1735, we now know that
∞
X
1
π2
=
n2
6
n=1
But why is this true? We can thank calculus for these results.
1
Contents
1 Introduction
3
2 Convergence
3
3 Euler’s First Attempt
4
4 Fourier Series
6
5 The Proof
6
6 References
7
2
1
Introduction
Pietro Mengoli asked the mathematical community in 1644 to find the sum
∞
X
1
1
1
1
= 2 + 2 + 2 + ...
2
n
1
2
3
n=1
After puzzling the cleverest of mathematicians, no results showed up. Fortunately, word got around to twenty-eight-year-old Leonhard Euler, whose solution
debuted him to the mathematical genius as he is know today. Euler has proven
2
that this sum converges to π6 . This problem was an important conquest in
number theory, as it relates to the Riemann-zeta function:
ζ (n) =
∞
X
1
1
1
1
= n + n + n + ...
n
i
1
2
3
i=1
Finding the complex zeros to this function is currently a $1-million Millenium
Prize Problem. But all of this was kickstarted by Leonhard Euler’s solution to
the Basel problem.
2
Convergence
2
We will define the Basel Problem sum to be β, and prove it to equal π6 . By
definition, β is a p-series with a value of n = 2. We can tell that it converges
right off the bat because 2 > 1. But, we want a bit more rigorous proof of its
convergence.
We are going to use the comparison test to prove its convergence. Term by
term,
1
1
1
1
1
1
1
+ 2 + 2 + 2 + 2 + 2 + 2 + ··· <
12
2
3
4
5
6
7
1
1
1
1
1
1
1
+ 2 + 2 + 2 + 2 + 2 + 2 + ··· <
12
2
2
4
4
4
4
1 1 1
+ + + ...
1 2 4
By comparison to the second series, which converges to 2 as a geometric
series with initial term 1 and common ratio of −1 < 12 < 1. So, by comparison,
we can see that β converges to some 0 < β < 2. A graph of the partial sums
helps visualize the convergence:
π2
6
•
•
•
•
•
3
•
•
•
•
•
Alternatively, β’s convergence can be proven by the integral test. Consider
the function
1
x2
One of the convergent integrals of this function is
∞
Z ∞
1
dx
= −
=1
x2
x 1
1
f (x) =
Using a right-endpoint Riemann sum with equal widths of 1 at each integral
value of x, we have
Z ∞
1
1
1
dx
≈ 2 + 2 + 2 + ...
x2
2
3
4
1
But, since f (x) is decreasing and concave up on the interval (1, ∞), the
approximation, and an underapproximation (shown in the figure below). Thus,
1
1
1
+ 2 + 2 + ··· < 1
2
2
3
4
This implies that the sequence is convergent and
0<
1<
1
1
1
1
+ 2 + 2 + 2 + ··· < 2
2
1
2
3
4
1
0.75
0.5
0.25
1
3
2
3
4
5
Euler’s First Attempt
Euler begins his research by giving a clever approximation of an integral that
yields β:
1
1
1
+ 2 + 2 + · · · = lim
2
n→∞
1
2
3
Z
1
0
Z
x
0
1 − yn
dydx ≈ 1.644924
1−y
2
Euler, being a clever mathematician, recognized this to be near π6 = 1.644934.
Using this information, he turned his attention to trigonometric functions, and
his “proof” is as follows.
4
Lemma 1.
sin x
x
is equivalent to its infinite product.
Proof. sin x has roots at 0, ±π, ±2π, ±3π, . . . . This means that sinx x has roots
at ±π, ±2π, ±3π, . . . . Leonhard Euler thus asserts without proof that
x
x
x
x
sin x
1+
1−
1+
··· =
= 1−
x
π
π
2π
2π
x2
x2
x2
1− 2
1 − 2 ...
1− 2
π
4π
9π
This is the downfall of his proof.
Lemma 2.
sin x
x
is equivalent to its infinite series.
Proof. Consider the Maclaurin series for
sin x = x −
x3
x5
+
− ...
3!
5!
Thus,
x2
x4
sin x
=1−
+
− ...
x
3!
5!
Lemma 3. β =
π2
6
Proof. From Lemmas 1 and 2, we can tell that
x2
x2
x2
x2
x4
1− 2
1− 2
1 − 2 ··· = 1 −
+
− ...
π
4π
9π
3!
5!
This means that we can end up expanding the left-hand equation and get
the same infinite series with the same coefficients. He proceeds to assert that,
using the coefficient of the x2 terms in the right-hand equation and after infinite
expansion of the left-hand side for just the x2 coefficient,
1
1
1
1
1
− = − 2 − 2 − 2 − ··· = − 2
3!
π
4π
9π
π
Solving for β, we get β =
1 1 1
+ + + ...
1 4 9
=−
β
π2
π2
6 .
But this proof is filled with many gaps, so a stronger and more rigorous
proof was sought.
5
4
Fourier Series
As with Taylor series expansions of functions as power series, there is a way of
representing functions in a different way using sin x and cos x called harmonic
analysis. The basis of harmonic analysis is the Fourier series. The Fourier
series says that a periodic function can be represented as an infinite sum in the
following way for some sequences of coefficients an and bn :
∞
f (x) =
a0 X
(an cos (nx) + bn sin (nx))
+
2
n=1
It turns out that these coefficients can be computed using these formulas:
Z
1 π
a0 =
f (x) dx
π −π
Z
1 π
an =
f (x) cos (nx) dx
π −π
Z
1 π
bn =
f (x) sin (nx) dx
π −π
Finding the Fourier series for specific functions comes in handy for this very
problem.
5
The Proof
Theorem 1. β =
π2
6
Proof. Using Fourier series, we can solve this problem rather easily. Euler pre2
dicted that β = π6 , so we know that the function somewhere will involve the
function f (x) = x2 . So let’s expand x2 using Fourier series.
π
Z
2π 2
1 x3
1 π 2
=
x dx =
a0 =
π −π
π 3 −π
3
Z π
1
bn =
x2 sin (nx) dx = 0 since x2 sin (nx) is odd
π −π
R
Now to find x2 cos (nx) dx, we will need to integrate by parts twice.
Z
Z
sin (nx)
x2 sin (nx)
− 2x
dx =
x2 cos (nx) dx =
n
n
Z
x2 sin (nx)
2
−
x sin (nx) dx =
n
n
Z
− cos (nx)
2 −x cos x
x2 sin (nx)
−
−
dx =
n
n
n
n
6
x2 sin (nx) 2x cos (nx) 2 sin (nx) dx
+
−
+C
n
n2
n3
Now, we must evaluate this as a definite integral. From −π to π, we compute
the integral to be
1
an =
π
Z
2 π 2 n2 − 2 sin (nπ) + 4nπ cos (nπ)
x cos (nx) dx =
n3 π
−π
π
2
But since the series only requires integral values of n, sin (nπ) = 0 for integral
n
n, and cos (nπ) = (−1) for integral values of n,
an =
4 (−1)
n2
n
This means that the Fourier series for x2 is
n
∞
x2 =
π 2 X 4 (−1)
+
cos (nx)
3
n2
n=1
Lastly, substitute x = π and end up with
∞
n
∞
π 2 X 4 (−1)
π 2 X 4 (−1)
+
cos
(nπ)
=
+
π =
3
n2
3
n2
n=1
n=1
2
2n
cos (nπ) =
∞
X
1
π2
π2
+4
=
+ 4β
3
n2
3
n=1
Solving for β, we get that β =
π2
6 .
This proof, much more substantial and rigorous than Euler’s proof, gives
valid evidence that
β=
6
∞
X
π2
1
=
2
n
6
n=1
References
• Sandifer, Ed. “How Euler Did It: Estimating the Basel Problem.” MAA.org.
Mathematical Association of America, Dec. 2003. Web. 4 June 2013.
• Sangwin, C.J. “An Infinite Series of Surprises.” Plus.Maths.org. Plus
Magazine, 1 Dec. 2001. Web. 4 June 2013.
7
Cody Johnson
