This article examines the famous Basel problem.


The Basel problem was the problem to evaluate the sum

$ \sum_{k=1}^\infty\dfrac{1}{k^2} $

in other words, $ \zeta(2) $ . First, we will see why it must converge here, using the integral test: If the series converge, the integral of the general term must converge. Let's test:

$ \lim_{x\to\infty}\int x^{-2}dx=\lim_{x\to\infty}\dfrac{x^{-1}}{-1}+C=\lim_{x\to\infty}-\frac{1}{x}+C=C $

Hence the series converge. The real problem is, to what number?

Euler's approach

Leonhard Euler was the first one to develop a clever approach to "solve" the problem, though not as rigorously as required. We will first demonstrate this approach and then give a rigorous proof.

The power series for the sine function is

$ \sin(x)=\sum_{k=0}^\infty(-1)^k\dfrac{x^{2k+1}}{(2k+1)!} $

so this holds:

$ \frac{\sin(x)}{x}=\sum_{k=0}^\infty(-1)^k\dfrac{x^{2k}}{(2k+1)!} $

which we can express more clearly as

$ 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots $

It is obvious that the coefficient of $ x^2 $ is 1/6. We will now obtain this coefficient through a different method. Obviously, the LHS function has zeros at every integer that is a multiple of pi, except 0. Through polynomial factoring, we obtain that we can show this as

$ (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots $

Using the difference of two squares formula, we simplify this to

$ (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots $

Rearranging factors, we get

$ \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 $

Now, with some simple algebra we obtain that the coefficient of $ x^2 $ in this expression is

$ \sum_{k=1}^\infty\dfrac{1}{(k\pi)^2} $

Since if two polynomials are equal, all coefficients must be equal; and this one is equal to the power series expansion, we get

$ \sum_{k=1}^\infty\dfrac{1}{(k\pi)^2}=\frac16 $

Multiplying thoroughly by $ \pi^2 $ , we obtain the famous solution to the Basel problem:

$ \sum_{k=1}^\infty\dfrac{1}{k^2}=\frac{\pi^2}{6} $

However, as we stated, this proof is not rigorous enough, whilst it is still true: The result is correct.

A rigorous proof

Now, since we know the answer, we are ready to give a rigorous proof. Using Fourier analysis, we know that for a function that is periodic over the interval $ (-\pi,\pi) $ with period $ 2\pi $ , the Fourier series are (please note that the function must be continous on the open interval and the left hand side aswell as the right hand side derivates must exist and be finite at every point of the open interval if we want equality between the function and its fourier series. If we, in additional, want equality to hold between the function and the fourier series on the whole interval $ [-\pi,\pi] $ then it must also hold that $ f(\pi)=f(-\pi) $

$ f(x)=\dfrac{a_0}{2}+\sum_{r=1}^\infty a_r\cos(rx)+\sum_{r=1}^\infty b_r\sin(rx) $


$ \begin{align} a_0&=\dfrac{1}{\pi}\int\limits_{-\pi}^\pi f(x)dx\\ a_r&=\dfrac{1}{\pi}\int\limits_{-\pi}^\pi f(x)\cos(rx)dx\\ b_r&=\dfrac{1}{\pi}\int\limits_{-\pi}^\pi f(x)\sin(rx)dx \end{align} $

We start out by finding the Fourier series for $ x^{2n} $ :

$ a_0=\dfrac{1}{\pi}\int\limits_{-\pi}^\pi x^{2n}dx=\frac{2\pi^{2n}}{2n+1} $

Since the function is even, the sine coefficient will vanish. Hence we just need to evaluate

$ a_r=\dfrac{1}{\pi}\int\limits_{-\pi}^\pi x^{2n}\cos(rx)dx $

First, we will evaluate the indefinite integral. If

$ F(n)=\int x^{2n}\cos(rx)dx $

then, we obtain the functional equation

$ F(n)=\frac{x^{2n}}{r}\sin(rx)+\frac{2n}{r^2}x^{2n-1}\cos(rx)-\frac{2n(2n-1)}{r^2}F(n-1) $

via integration by parts. By induction or simple observation, this functional equation yields the following formula:

$ F(n)=\sin(rx)\sum_{k=0}^n\dfrac{(-1)^{k}(2n)!x^{2n-2k}}{(2n-2k)!r^{2k+1}}+\cos(rx)\sum_{k=0}^{n-1}\frac{(-1)^k(2n)!x^{2n-2k-1}}{(2n-2k-1)!r^{2k+2}} $

The sine term vanishes in the Fourier series because we will use integer multiples of pi in the sine function, and that yields zero, and the cosine function yields $ (-1)^r $ . Hence, we are left with

$ G_n(r)=\int\limits_{-\pi}^\pi x^{2n}\cos(rx)dx=2(-1)^r\sum_{k=0}^{n-1}\dfrac{(-1)^k(2n)!\pi^{2n-2k-1}}{(2n-2k-1)!r^{2k+2}} $

Substituting these into our original Fourier series, we get

$ x^{2n}=\dfrac{\pi^{2n}}{2n+1}+\frac{1}{\pi}\sum_{r=1}^\infty G_{n}(r)\cos(rx) $

We now put $ n=1 $ and $ x=\pi $ to obtain

$ \pi^2=\dfrac{\pi^2}{3}+\frac{1}{\pi}\sum_{r=1}^\infty G_1(r)(-1)^r $

The general term of the sum simplifies to

$ G_1(r)(-1)^r=2(-1)^{2r}\sum_{k=0}^0\dfrac{(-1)^k2!\pi^{1-2k}}{(1-2k)!r^{2k+2}}=\frac{4\pi}{r^2} $

Substituting back:

$ \pi^2=\dfrac{\pi^2}{3}+4\sum_{r=1}^\infty\frac{1}{r^2} $

The value we want to evaluate just appeared. We set it to C:

$ \pi^2=\frac{\pi^2}{3}+4C $

Solving this equation:

$ \begin{align} &3\pi^2=\pi^2+12C\\&2\pi^2=12C\\&\dfrac{\pi^2}{6}=C=\sum_{r=1}^\infty\frac{1}{r^2} \end{align} $

and this completes the rigorous proof.