# Basel Problem

## Theorem

$\displaystyle \zeta \left({2}\right) = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.

## Proof 1

$\displaystyle \frac 1 {n^2} = \int_0^1 \! \int_0^1 (xy)^{n-1} \ \mathrm dx \ \mathrm dy$

Hence:

$\displaystyle \sum_{n \mathop = 1}^\infty {\frac 1 {n^2}} = \sum_{n = 1}^\infty \int_0^1 \! \int_0^1 (xy)^{n-1} \ \mathrm dx \ \mathrm dy$

By the Monotone Convergence Theorem (Real Analysis) we have:

$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \int_0^1 \! \int_0^1 \left({\sum_{n \mathop = 1}^\infty \left({x y}\right)^{n-1} }\right) \mathrm dx \ \mathrm dy = \int_0^1 \! \int_0^1 \frac {\mathrm dx \ \mathrm dy} {1-xy}$

Let $\displaystyle (u,v) = \left({\frac{x+y} 2, \frac{y-x} 2}\right)$ so that $(x,y) = (u-v, \ u+v)$.

Then:

$\displaystyle \zeta \left({2}\right) = 2 \iint_S \frac{\mathrm du \ \mathrm dv} {1 - u^2 + v^2}$

where $S$ is the square defined by the coordinates $\displaystyle (0,0), \ \left({\frac 1 2, -\frac 1 2}\right), \ (1,0), \ \left({\frac 1 2, \frac 1 2}\right)$.

Exploiting the symmetry of the square we have:

$\displaystyle \zeta \left({2}\right) = 4 \left({\int_0^{\frac 1 2} \! \int_0^u \frac {\mathrm dv \ \mathrm du} {1-u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1-u} \frac {\mathrm dv \ \mathrm du} {1 - u^2 + v^2}}\right)$

After some straightforward calculations we get:

$\displaystyle \zeta \left({2}\right) = 4 \left( \int_{0}^{\frac{1}{2}}{\frac{\arcsin u}{\sqrt{1-u^{2}}}\,\mathrm du} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt{1-u^2}} \left( \frac{\pi } 4-\frac{\arcsin u} 2 \right)\,\mathrm du} \right) = 4 \left( \frac{\pi^2}{72}+\frac{\pi^2} {36} \right) = \frac {\pi^2} 6$

$\blacksquare$

## Proof 2

Let

$\displaystyle f(x) = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots$.

We note that

$\displaystyle f(0)=1-\frac{0}{3!}+\frac{0}{5!}-\frac{0}{7!}+\frac{0}{9!}-\cdots = 1$.

Now for all $x\neq 0$ $\frac{x}{x}=1$. Therefore,

$\displaystyle f(x) = \frac{x}{x}f(x)=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots}{x}$.

From the analytic definition of sine:

$\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$.

Therefore, for all $x\neq 0$

$\displaystyle f(x)=\frac{\sin x}{x}$.

From Euler's formula for the sine function we know that

$\displaystyle f(x) = \left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right) \left( 1 - \frac{x}{2\pi} \right) \left( 1 + \frac{x}{2\pi} \right) \left( 1 - \frac{x}{3\pi} \right) \left( 1 + \frac{x}{3\pi} \right) \cdots = \left[ 1 - \frac{x^2}{\pi^2} \right] \left[ 1 - \frac{x^2}{4\pi^2} \right] \left[ 1 - \frac{x^2}{9\pi^2} \right] \left[ 1 - \frac{x^2}{16\pi^2} \right]\cdots$.

Expanding this infinite product we get

$\displaystyle \left[ 1 - \frac{x^2}{\pi^2} \right] \left[ 1 - \frac{x^2}{4\pi^2} \right] \left[ 1 - \frac{x^2}{9\pi^2} \right] \left[ 1 - \frac{x^2}{16\pi^2} \right]\cdots = 1 - \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 + \left(\dots\right)x^4-\cdots$.

From this

$\displaystyle 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots = 1 - \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 + \left(\dots\right)x^4-\cdots$.

Now the first element in each infinite sum is equal to the first element in the second and so on. From this we know that

$\displaystyle \frac{x^2}{3!} = \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2$.

Dividing each side of this equation by $x^2$ and factoring out $\frac{1}{\pi^2}$ on the right side of the equation we obtain

$\displaystyle \frac{1}{6} = \frac{1}{\pi^2}\left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \right)$.

Multiplying each side of this equation by $\pi^2$ we obtain

$\displaystyle \frac{\pi^2}{6} = \left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \right)$.

Therefore,

$\displaystyle \zeta \left({2}\right) = \sum\limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$.

$\blacksquare$

## Proof 3

Let $x \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$ and let $n$ be a non-negative integer.

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \frac {\cos \left({2n + 1}\right) x + i \sin \left({2n + 1}\right) x} {\sin^{2n + 1} x}$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\left({\cos x + i \sin x}\right)^{2n + 1} } {\sin^{2n + 1} x}$$ $$\displaystyle$$ $$\displaystyle$$ De Moivre's Formula $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({\cot x + i}\right)^{2n + 1}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{r \mathop = 0}^{2n + 1} \binom{2n + 1}{r} i^r \cot^{2n + 1 - r} x$$ $$\displaystyle$$ $$\displaystyle$$ Binomial Theorem $$\displaystyle$$ $$\displaystyle \implies$$ $$\displaystyle$$ $$\displaystyle \frac {\sin \left({2n + 1}\right)x} {\sin^{2n + 1} x}$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} (-1)^r \cot^{2 \left({n - r}\right)} x$$ $$\displaystyle$$ $$\displaystyle$$ equating imaginary parts

Let $x_k = \dfrac {k \pi} {2n + 1}$ for $k = 1, 2, \ldots, n$.

Then $\sin \left({2n + 1}\right) x_k = 0$.

So we have:

$\displaystyle \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} (-1)^r \cot^{2 \left({n - r}\right)} x_k = 0$

for $k = 1, 2, \ldots, n$.

The numbers $x_k$ are all distinct and in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

By Shape of Cotangent Function, $\cot x$ is positive and injective in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$

Therefore $\cot^2 x$ is also injective in this interval.

Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$.

These numbers are the $n$ distinct roots of the $n$th degree polynomial:

$\displaystyle f(c) := \sum_{r \mathop = 0}^n \binom {2n + 1} {2r + 1} (-1)^r c^{n - r}$

By Viète's Formulas, we can calculate the sum of the roots:

$\displaystyle \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2n + 1} 3} {\binom {2n + 1} 1} = \frac {2n \left({2n - 1}\right)} 6$

Using the identity $\cot^2 x = \csc^2 x - 1$ we can similarly deduce:

$\displaystyle \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2n \left({2n - 1}\right)} 6 + n = \frac {2n \left({2n + 2}\right)} 6$

By Shape of Sine Function, $\sin x$ is positive in the interval $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

So from the inequality $\sin x < x < \tan x$ for $x \in \left({0 \,.\,.\, \dfrac \pi 2}\right)$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval.

Summing this inequality from $x_1$ to $x_n$ gives:

$\displaystyle \frac {2n \left({2n - 1}\right)} 6 < \sum_{k \mathop = 1}^n \left({\frac {2n + 1} {k \pi}}\right)^2 < \frac {2n \left({2n + 2}\right)} 6$

or equivalently:

$\displaystyle \frac {\pi^2} 6 \frac {2n \left({2n - 1}\right)} {\left({2n + 1}\right)^2} < \sum_{k \mathop = 1}^n \frac 1 {k^2} < \frac {\pi^2} 6 \frac {2n \left({2n + 2}\right)} {\left({2n + 1}\right)^2}$

By Combination Theorem for Limits of Functions, the left and right hand sides tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity.

Therefore by Squeeze Theorem:

$\displaystyle \zeta \left({2}\right) = \sum_{k \mathop = 1}^\infty \frac 1 {k^2} = \frac {\pi^2} 6$

$\blacksquare$

## Historical Note

Although often credited to Euler, it has been suggested that it was in fact solved by Nicolaus I Bernoulli.