Basel Problem

The Basel problem asks for the precise summation of the reciprocals of the squares of positive integers, i.e. the precise sum of the infinite series:

\[\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } = } \lim _{ n\rightarrow \infty }{ \left( \frac { 1 }{ 1^{ 2 } } +\frac { 1 }{ 2^{ 2 } } +\frac { 1 }{ 3^{ 2 } } +\cdots +\frac { 1 }{ { n }^{ 2 } } \right) . }\]

It asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be \(\frac{\pi^2}{6}\) and announced this discovery in 1735.

Basel Theorem

We have

\[\sin { x } =x-\frac { { x }^{ 3 } }{ 3! } +\frac { { x }^{ 5 } }{ 5! } -\frac { { x }^{ 7 } }{ 7! } +\cdots.\]

Dividing by \(x\), we have

\[\frac { \sin { x } }{ x } = 1 - \frac { { x }^{ 2 } }{ 3! } + \frac { { x }^{ 4 } }{ 5! } -\frac { { x }^{ 6 } }{ 7! } +\cdots.\]

Using the Weierstrass factorization theorem, it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials:

\[\begin{align} \frac { \sin { x } }{ x } &= \left( 1-\frac { x }{ \pi } \right) \left( 1+\frac { x }{ \pi } \right) \left( 1-\frac { x }{ 2\pi } \right) \left( 1+\frac { x }{ 2\pi } \right) \left( 1-\frac { x }{ 3\pi } \right) \left( 1+\frac { x }{ 3\pi } \right) \ldots\\ &=\left( 1-\frac { { x }^{ 2 } }{ { \pi }^{ 2 } } \right) \left( 1-\frac { { x }^{ 2 } }{ { 4\pi }^{ 2 } } \right) \left( 1-\frac { { x }^{ 2 } }{ { 9\pi }^{ 2 } } \right) \ldots. \end{align}\]

If we formally multiply out this product and collect all the \(x^2\) terms (we are allowed to do so because of Newton's identities), we see that the \(x^2\) coefficient of \(\frac{\sin{x}}{x}\) is

\[-\left( \frac { 1 }{ { \pi }^{ 2 } } +\frac { 1 }{ { 4\pi }^{ 2 } } +\frac { 1 }{ { 9\pi }^{ 2 } } ... \right) =\frac { -1 }{ { \pi }^{ 2 } } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }. \]

But from the original infinite series expansion of \(\frac{\sin{x}}{x}\), the coefficient of \(x^2\) is \(-\frac {1}{6} \). These two coefficients must be equal, so

\[-\frac { 1 }{ 6 } =\frac { -1 }{ { \pi }^{ 2 } } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }. \]

Multiplying through both sides of this equation by \(-\pi^2\) gives the sum of the reciprocals of the positive square integers:

\[ \sum _{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}. \ _\square \]