The Basel Problem: Euler’s Triumph and the Birth of Modern Analysis
The Problem Statement
In 1650, Italian mathematician Pietro Mengoli posed a deceptively simple question that would stump the greatest mathematical minds for nearly a century:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = ?$$
The series clearly converges (as Jakob Bernoulli proved in 1689), and numerical approximation yields approximately $1.644934\ldots$, but what is the exact value?
This became known as the Basel Problem, named after the Swiss city that was home to both Leonhard Euler and the Bernoulli family.
Historical Context
The Failed Attempts
Before Euler’s breakthrough, the problem defeated formidable intellects:
- Jakob Bernoulli (1654-1705): Proved convergence, bounded the sum between 1.5 and 2
- Johann Bernoulli (1667-1748): Tried multiple approaches, all unsuccessful
- Gottfried Wilhelm Leibniz: Applied his calculus techniques, no success
- James Stirling and Abraham de Moivre: Both attempted and failed
The difficulty was not just finding an approximation, but determining the exact closed form - and proving it rigorously.
Euler’s Moment of Glory
In 1735, at age 28, Leonhard Euler announced his solution:
$$\boxed{\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}}$$
This unexpected connection between the discrete world of integers and the transcendental constant $\pi$ was revolutionary. Euler’s solution brought him immediate fame throughout the mathematical community.
On the day his brother Jakob died, Johann Bernoulli reportedly said: “If only my brother were alive now.”
Proof 1: Euler’s Original Approach (1735)
Euler’s first proof, while not rigorous by modern standards, demonstrated brilliant intuition that would later be vindicated.
The Taylor Series Foundation
We begin with the Maclaurin series for $\sin x$:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
Dividing both sides by $x$:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots \quad \quad (1)$$
The Product Representation
Now comes Euler’s bold leap. The function $\sin x$ has zeros at $x = 0, \pm \pi, \pm 2\pi, \pm 3\pi, \ldots$
For $\frac{\sin x}{x}$, the zeros (excluding the removable singularity at $x=0$) are at $x = \pm \pi, \pm 2\pi, \pm 3\pi, \ldots$
Euler reasoned that, like a polynomial, this function could be factored over its roots:
$$\frac{\sin x}{x} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots$$
Pairing the factors:
$$\frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots$$
Therefore:
$$\frac{\sin x}{x} = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2\pi^2}\right) \quad \quad (2)$$
Extracting the Coefficient
Expanding the infinite product (2):
$$\frac{\sin x}{x} = 1 - \left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots\right)x^2 + \text{(higher order terms)}$$
$$\frac{\sin x}{x} = 1 - \frac{1}{\pi^2}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}\right)x^2 + \text{(higher order terms)} \quad \quad (3)$$
Comparing the coefficient of $x^2$ in equations (1) and (3):
From (1): coefficient is $-\frac{1}{6}$
From (3): coefficient is $-\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}$
Therefore:
$$-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}$$
Solving for the sum:
$$\boxed{\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}}$$
Euler’s Leap and Later Validation
Euler’s argument involved several unjustified steps by the standards of 1735:
- Can an infinite product be treated like a polynomial factorization?
- Can we expand an infinite product and compare coefficients term-by-term?
- Does the series representation converge uniformly?
It took over 100 years and the development of complex analysis (particularly Weierstrass’s theory of entire functions) to rigorously justify Euler’s approach. The key result is the Weierstrass factorization theorem:
Theorem (Weierstrass): Every entire function can be represented as a product involving its zeros.
For $\sin(\pi z)$, this gives:
$$\sin(\pi z) = \pi z \prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$$
This validates Euler’s factorization exactly!
Proof 2: Rigorous Proof via Fourier Analysis
We now present a completely rigorous proof using Parseval’s identity from Fourier analysis.
Fourier Series Preliminaries
Definition: For a function $f \in L^2[-\pi, \pi]$, the Fourier series is:
$$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos(nx) + b_n\sin(nx)\right)$$
where the Fourier coefficients are:
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx),dx$$
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx),dx$$
Parseval’s Identity
Theorem (Parseval’s Identity): If $f \in L^2[-\pi,\pi]$ with Fourier coefficients $a_n, b_n$, then:
$$\frac{1}{\pi}\int_{-\pi}^{\pi}|f(x)|^2,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2 + b_n^2)$$
This is the $L^2$ analogue of the Pythagorean theorem - the “energy” in the function equals the sum of “energies” in its frequency components.
Application to $f(x) = x$
Consider $f(x) = x$ on $[-\pi, \pi]$.
Computing $a_n$:
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\cos(nx),dx$$
Since $x\cos(nx)$ is an odd function:
$$a_n = 0 \quad \text{for all } n \geq 0$$
Computing $b_n$:
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx),dx$$
Using integration by parts with $u = x$, $dv = \sin(nx),dx$:
$$b_n = \frac{1}{\pi}\left[\left.-\frac{x\cos(nx)}{n}\right|{-\pi}^{\pi} + \frac{1}{n}\int{-\pi}^{\pi}\cos(nx),dx\right]$$
The integral of $\cos(nx)$ over a full period vanishes:
$$b_n = \frac{1}{\pi}\left[-\frac{\pi\cos(n\pi)}{n} + \frac{\pi\cos(-n\pi)}{n}\right]$$
$$b_n = \frac{1}{\pi} \cdot \frac{-2\pi\cos(n\pi)}{n} = \frac{-2\cos(n\pi)}{n} = \frac{2(-1)^{n+1}}{n}$$
Therefore, the Fourier series for $f(x) = x$ is:
$$x \sim 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx)$$
Applying Parseval’s Identity
The left-hand side:
$$\frac{1}{\pi}\int_{-\pi}^{\pi}x^2,dx = \frac{1}{\pi} \cdot \frac{2\pi^3}{3} = \frac{2\pi^2}{3}$$
The right-hand side (with $a_n = 0$ for all $n$):
$$\sum_{n=1}^{\infty}b_n^2 = \sum_{n=1}^{\infty}\left(\frac{2(-1)^{n+1}}{n}\right)^2 = 4\sum_{n=1}^{\infty}\frac{1}{n^2}$$
By Parseval’s identity:
$$\frac{2\pi^2}{3} = 4\sum_{n=1}^{\infty}\frac{1}{n^2}$$
Therefore:
$$\boxed{\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}}$$
Why This Proof is Rigorous
This proof relies only on:
- Integration by parts (elementary calculus)
- Parseval’s identity (proven rigorously in functional analysis)
- The convergence of Fourier series for $f(x) = x$ (well-established)
No questionable manipulations of infinite products are required.
Proof 3: Complex Analysis via Residues
For completeness, we sketch a proof using the calculus of residues.
The Cotangent Function
Consider the function:
$$g(z) = \frac{\pi \cot(\pi z)}{z^2}$$
This function has:
- A pole of order 2 at $z = 0$
- Simple poles at $z = n$ for all $n \in \mathbb{Z} \setminus {0}$
Residue Calculation
The residue at $z = n$ (for $n \neq 0$) is:
$$\text{Res}{z=n}(g) = \lim{z \to n}(z-n)\frac{\pi\cot(\pi z)}{z^2} = \frac{1}{n^2}$$
Contour Integration
Consider a square contour $C_N$ with vertices at $(\pm N \pm Ni)\pi$ where $N$ is a half-integer.
By the residue theorem:
$$\oint_{C_N}g(z),dz = 2\pi i \sum_{\text{poles inside}} \text{Res}$$
As $N \to \infty$, the integral over $C_N$ vanishes (which can be proven by bounding $|\cot(\pi z)|$), and we obtain:
$$0 = 2\pi i \left(\text{Res}{z=0} + 2\sum{n=1}^{\infty}\frac{1}{n^2}\right)$$
The residue at $z=0$ can be computed via Laurent expansion:
$$\text{Res}_{z=0}(g) = -\frac{\pi^2}{3}$$
Therefore:
$$0 = 2\pi i\left(-\frac{\pi^2}{3} + 2\sum_{n=1}^{\infty}\frac{1}{n^2}\right)$$
$$\boxed{\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}}$$
Connection to the Riemann Zeta Function
The Basel problem is the special case $s=2$ of the Riemann zeta function:
$$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} \quad \text{for } \Re(s) > 1$$
Thus:
$$\zeta(2) = \frac{\pi^2}{6}$$
Euler’s Generalizations
Euler didn’t stop at $\zeta(2)$. He computed:
$$\zeta(4) = \sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}$$
$$\zeta(6) = \sum_{n=1}^{\infty}\frac{1}{n^6} = \frac{\pi^6}{945}$$
$$\zeta(8) = \sum_{n=1}^{\infty}\frac{1}{n^8} = \frac{\pi^8}{9450}$$
In general, for even integers, Euler found:
$$\zeta(2k) = \frac{(-1)^{k+1}(2\pi)^{2k}B_{2k}}{2(2k)!}$$
where $B_{2k}$ are the Bernoulli numbers.
The Odd Zeta Values Mystery
Remarkably, we still don’t have closed forms for odd values! For instance:
$$\zeta(3) = \sum_{n=1}^{\infty}\frac{1}{n^3} = 1.202056903159594\ldots$$
This is Apéry’s constant, proven irrational by Roger Apéry in 1979, but we don’t know if it can be expressed in terms of $\pi$ and other known constants.
The odd zeta values remain one of the deepest mysteries in number theory.
Euler’s Product Formula
Euler discovered a profound connection between $\zeta(s)$ and prime numbers:
$$\zeta(s) = \prod_{p \text{ prime}}\frac{1}{1-p^{-s}}$$
For $s=2$:
$$\frac{\pi^2}{6} = \prod_{p \text{ prime}}\frac{1}{1-p^{-2}} = \frac{1}{1-2^{-2}} \cdot \frac{1}{1-3^{-2}} \cdot \frac{1}{1-5^{-2}} \cdots$$
$$\frac{\pi^2}{6} = \frac{4}{3} \cdot \frac{9}{8} \cdot \frac{25}{24} \cdot \frac{49}{48} \cdots$$
This equation connects:
- The transcendental number $\pi$
- All natural numbers (via $\sum 1/n^2$)
- All prime numbers (via the product)
It’s a statement of breathtaking unity in mathematics.
Probability Interpretation
The Basel problem has a beautiful probabilistic interpretation:
Theorem: The probability that two randomly chosen positive integers are coprime (share no common factors) is:
$$P(\gcd(m,n)=1) = \frac{6}{\pi^2} \approx 0.6079$$
Proof Sketch: Using Euler’s product:
$$\frac{1}{\zeta(2)} = \prod_{p}\left(1-\frac{1}{p^2}\right)$$
The probability that $p$ does not divide both $m$ and $n$ is $1-1/p^2$. For all primes:
$$P(\gcd(m,n)=1) = \prod_p\left(1-\frac{1}{p^2}\right) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}$$
Thus, $\pi$ appears in a purely discrete probability problem!
Modern Significance
Foundation for Riemann Hypothesis
Euler’s work directly inspired Riemann’s 1859 paper extending $\zeta(s)$ to complex values. The Riemann Hypothesis - the most important unsolved problem in mathematics - concerns the zeros of $\zeta(s)$ in the complex plane.
Applications
The techniques developed to solve the Basel problem are now fundamental to:
- Analytic Number Theory: Distribution of primes
- Quantum Field Theory: Regularization and renormalization
- Statistical Mechanics: Partition functions
- Random Matrix Theory: Eigenvalue distributions
- Fourier Analysis: Signal processing and quantum mechanics
Beautiful Consequences
Sum of Odd Reciprocal Squares
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{8}$$
Proof:
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$
$$\frac{\pi^2}{6} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$
$$\frac{\pi^2}{6} = \frac{\pi^2}{24} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{8}$$
Alternating Series
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}$$
This follows from considering the Fourier series of $f(x) = x$ on $[0, \pi]$.
Reflection on Mathematical Beauty
The Basel problem exemplifies mathematical beauty through:
-
Unexpected Connections: Why should summing reciprocal squares involve $\pi$?
-
Multiple Pathways: The problem can be solved via analysis, algebra, complex analysis, probability, and Fourier theory - each revealing different insights.
-
Generative Power: Solving this one problem opened entirely new fields (Riemann zeta function, distribution of primes, modern analysis).
-
Elegance: The answer $\pi^2/6$ is shockingly simple given the infinite complexity of the sum.
-
Historical Drama: 90 years of failure by giants, then breakthrough by a 28-year-old.
As mathematician William Dunham wrote: “Euler’s solution to the Basel problem is one of those little gems that brighten the history of mathematics.”
Computational Verification
Modern computation confirms:
$$\sum_{n=1}^{10^6}\frac{1}{n^2} \approx 1.644933$$
$$\frac{\pi^2}{6} \approx 1.644934$$
The series converges slowly - to get $k$ decimal places requires roughly $10^k$ terms. But Euler’s formula gives infinite precision instantly.
Conclusion
The Basel problem stands as a monument to human mathematical achievement. What began as Pietro Mengoli’s curious question became:
- Euler’s pathway to fame
- Foundation for the Riemann zeta function
- Inspiration for modern analysis and number theory
- A bridge between discrete and continuous mathematics
- One of the most beautiful results in all of mathematics
The fact that:
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \cdots = \frac{\pi^2}{6}$$
continues to inspire wonder in mathematicians and students alike, nearly 300 years after Euler’s breakthrough.
Further Exploration
For those wanting to go deeper:
- Prove $\zeta(4) = \pi^4/90$ using similar methods
- Explore the functional equation of $\zeta(s)$
- Study Bernoulli numbers and their connection to zeta values
- Investigate why odd zeta values resist closed forms
- Learn about L-functions and modern generalizations
The Basel problem is not just a beautiful result - it’s a gateway to some of the deepest mathematics ever developed.