# Basel Problem/Proof 11

## Contents

## Theorem

- $\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.

## Proof

From the Euler Formula for Sine Function, we have:

- $\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$

Therefore, if we divide by x, we get

- $\displaystyle \dfrac {\sin x} {x} = \dfrac {x} {x} \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \paren {1 - \frac {x^2} {1^2 \pi^2} } x \paren {1 - \frac {x^2} {2^2 \pi^2} } x \paren {1 - \frac {x^2} {3^2 \pi^2} } x \cdots$

- $\displaystyle $

From the Power Series Expansion for Sine Function, we have:

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

Therefore, if we divide by x, we get

- $\displaystyle \dfrac {\sin x} {x} = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n + 1}\right)!} = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$

- $\displaystyle $

Equating the infinite product with the infinite sum, we have:

- $\displaystyle \dfrac {\sin x} {x} = \dfrac {x} {x} \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n + 1}\right)!}$

- $\displaystyle $

- $\displaystyle \dfrac {\sin x} {x} = \paren {1 - \frac {x^2} {1 \pi^2} } x \paren {1 - \frac {x^2} {4 \pi^2} } x \paren {1 - \frac {x^2} {9 \pi^2} } x \cdots = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$

- $\displaystyle $

From this, we can see the following relationships:

- $\displaystyle - \frac {x^2} {3!} = - \frac {x^2} { \pi^2} \paren {1 + \frac {1} {4} + \frac {1} {9} + \frac {1} {16} \cdots} ($"Choose 1" - used to calculate $ \map \zeta 2)$

- $\displaystyle $

- $\displaystyle + \frac {x^4} {5!} = +\frac {x^4} { \pi^4} \paren {\paren {1} \paren {\frac {1} {4}} + \paren {1} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {16}} + \cdots + \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots + \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots} ($"Choose 2" - used to calculate $ \map \zeta 4)$

- $\displaystyle $

- $\displaystyle - \frac {x^6} {7!} = - \frac {x^6} { \pi^6} \paren {\paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots + \paren {1} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots + \paren {\frac {1} {4}} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots} ($"Choose 3" - used to calculate $\map \zeta 6)$

- $\displaystyle $

The equations above can be used to calculate the zeta function at even integer arguments.

Notice that

"Choose $1$" is the summation of each term chosen once.

"Choose $2$" is the summation of the product of every non-repeating combination of two terms.

"Choose $3$" is the summation of the product of every non-repeating combination of three terms.

- $\displaystyle $

For example, from "Choose $1$" above:

- $\displaystyle \map \zeta 2 = \dfrac { \pi^2} {3!}$

$\blacksquare$

## Historical Note

The Basel Problem was first posed by Pietro Mengoli in $1644$.

Its solution is generally attributed to Leonhard Euler , who solved it in $1734$ and delivered a proof in $1735$.

However, it has also been suggested that it was in fact first solved by Nicolaus I Bernoulli.

Jacob Bernoulli had earlier established that the series was convergent, but had failed to work out what to.

The problem is named after Basel, the home town of Euler as well as of the Bernoulli family.

*If only my brother were alive now.*