Chris Henson


May 27, 2020
The Basel Problem

In the past two posts, I have mentioned the Basel problem of finding the solution to the series: $$ \sum _{n=1}^{\infty } \frac {1}{n^{2}} =\frac {1}{1^{2}} + \frac {1}{2^{2}} + \frac {1}{3^{2}} + \cdots $$

There are several methods of solving this problem. I will follow an argument given by the mathematician Leonhard Euler in his book Introductio in analysin infinitorum §167, written in 1748. Both Euler and this text are without match in the world of mathematics. In the words of Laplace: "Read Euler, read Euler, he is the master of us all."

As is typical of Euler, he solves this problem in a way that would not be considered rigorous by modern mathematical standards, yet still has much value and insight centuries after his death. I will outline this solution here. First, note that we can factor $a^n-z^n$ as: $$ \prod_{2k \leq n} \left( a^2 - 2az\cos\left(\frac{2k\pi}{n}\right)+z^2 \right), \quad k \in \mathbb{N} $$

Note that we must account for the case of factors that are perfect squares. If $k=0$, we have $a^2 - 2az + z^2$ which gives the factor $(a-z)$. If $n$ is even and $2k = n$, $a^2 + 2az + x^2$ gives the factor $(a+z)$

As an example, we have: $$ a^5-z^5 = (a-z)\left( a^2 - 2az\cos\left(\frac{2}{5}\pi\right)+z^2\right)\left( a^2 - 2az\cos\left(\frac{4}{5}\pi\right)+z^2\right) $$ Keep this in the mind, we'll use it a bit later!

Next Euler considers the general equation: $$ 1 + Az + Bz^2 + Cz^3 + \dots = (1+\alpha z)(1+\beta z)(1+\gamma z)\dots $$ which is where the bit of unrigorous intuition takes place. He assumes that in equating the Taylor series on the left-hand side with the infinite product on the right that he may manipulate them term by term. If you expand the product in this sense you obtain: \begin{align} A &= \alpha + \beta + \gamma + \dots \\ B &= \alpha\beta + \alpha\gamma + \dots + \beta\gamma + \dots\\ \vdots \end{align} where each of the Latin terms $A, B, C, \dots$ which is a coefficient of $z^n$ is the sum of all combinations of length n of Greek letters.

Finally, denote the sum of powers of Greek letters as: \begin{align} P &= \alpha + \beta + \gamma + \dots\\ Q &= \alpha^2 + \beta^2 + \gamma^2 + \dots\\ \vdots \end{align} (it is a little ambiguous that these are also Latin letters, but it is okay since we will only look at the first few of each sequence.)

Note that we can write these in terms of each other: \begin{align} P &= A\\ Q &= AP - 2B\\ R &= AQ - BP - 3C\\ \vdots \end{align}

Now we can circle back to the factor trick we used earlier with a special sort of series. Recall that definition of the exponential function is: $$ e^x = \lim_{j \to \infty} \left(1+{\frac {x}{j}}\right)^{j} $$ in typical Euler fashion, instead of a limit he manipulates $j$ as a value that is infinite. So we have: $$ e^x-e^{-x} = \left(1+{\frac {x}{j}}\right)^{j} - \left(1-{\frac {x}{j}}\right)^{j} $$ Notice that if we let $n=j$, $a = 1+\frac{x}{j}$, and $z = 1-\frac{x}{j}$ that we can use our factoring formula! We have our factors as: \begin{align} &\prod_{k=0}^\infty \left( \left(1+\frac{x}{j}\right)^2 - 2\left(1+\frac{x}{j}\right)\left(1-\frac{x}{j}\right) \cos\left(\frac{2k\pi}{j}\right) + \left(1-\frac{x}{j}\right)^2 \right) \\ =& \prod_{k=0}^\infty \left( 2 + 2\frac{x^2}{j^2} - 2\left( 1-\frac{x^2}{j^2}\right)\cos\left(\frac{2k\pi}{j}\right) \right)\\ =& \prod_{k=0}^\infty \left( \frac{4x^2}{j^2} + \frac{4k^2}{j^2}\pi^2 - \frac{4k^2\pi^2x^2}{j^4} \right)\\ =& \prod_{k=0}^\infty \left( \frac{4k^2\pi^2}{j^2}\left( 1+\frac{x^2}{k^2\pi^2} - \frac{x^2}{j^2}\right) \right)\\ =& \prod_{k=0}^\infty \left( 1+\frac{x^2}{k^2\pi^2}\right) \end{align} Here we have eliminated terms where $j$ being infinite causes a term to tend towards zero and used $\cos\left(\frac{2k\pi}{j}\right) = 1 - \frac{2k^2\pi^2}{j^2}$. In fact, with the exception that the first factor which will be $x$, we have derived an infinite product for hyperbolic sine.

Dividing by $2x$ we have: $$ \frac{\sinh(x)}{x} = \frac{e^x-e^{-x}}{2}\frac{1}{x} = \left(1+\frac{x^2}{\pi^2}\right)\left(1+\frac{x^2}{4\pi^2}\right)\left(1+\frac{x^2}{9\pi^2}\right)\dots = 1+\frac{x^2}{3!} + \frac{x^4}{5!} + \frac{x^6}{7!} + \dots $$ So setting $x^2 = \pi^2 z$ gives: $$ 1+\frac{\pi^2}{3!} + \frac{\pi^4}{5!} + \frac{\pi^6}{7!} + \dots = (1+z)\left(1+\frac{1}{4}z\right)\left(1+\frac{1}{9}z\right)\left(1+\frac{1}{16}z\right)\dots $$ And using the polynomial notation from earlier we have \begin{align} A &= \frac{\pi^2}{3!} = \frac{\pi^2}{6}\\ B &= \frac{\pi^2}{5!} = \frac{\pi^4}{120}\\ C &= \frac{\pi^2}{7!} = \frac{\pi^6}{5040}\\ \vdots\\ P &= 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots\\ Q &= 1 + \frac{1}{4^2} + \frac{1}{9^2} + \frac{1}{16^2} + \dots\\ R &= 1 + \frac{1}{4^3} + \frac{1}{9^3} + \frac{1}{16^3} + \dots\\ \vdots \end{align} and we finally have the desired: $$ \sum _{n=1}^{\infty } \frac {1}{n^{2}} =\frac {1}{1^{2}} +\frac {1}{2^{2}} + \frac {1}{3^{2}} +\cdots = \dfrac{\pi^2}{6}$$ In fact, using the same polynomial method gives us an entire family of these equation where we substitute the power 2 for another even integer: \begin{align} P = \sum _{n=1}^{\infty } \frac {1}{n^{2}} &=\frac {1}{1^{2}} +\frac {1}{2^{2}} + \frac {1}{3^{2}} +\cdots = \dfrac{\pi^2}{6}\\ Q = \sum _{n=1}^{\infty } \frac {1}{n^{4}} &=\frac {1}{1^{4}} +\frac {1}{2^{4}} + \frac {1}{3^{4}} +\cdots = \dfrac{\pi^4}{90}\\ R = \sum _{n=1}^{\infty } \frac {1}{n^{6}} &=\frac {1}{1^{6}} +\frac {1}{2^{6}} + \frac {1}{3^{6}} +\cdots = \dfrac{\pi^6}{945}\\ \vdots \end{align}

What an interesting bit of mathematics, as well as an interesting window into the mathematics of the 18th century. In fact, this series has much deeper mathematical connections and the following function: $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$ is equal to the Riemann zeta function for values of $s$ within the radius of convergence for the series. The complex-valued analytic continuation of this series has deep connections to prime numbers and even applications to statistics and physics. A subject for a later post!