# Basel Problem/Proof 2

## 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

Let:

- $\displaystyle P_k = x \prod_{n \mathop = 1}^k \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$

We note that:

\(\displaystyle P_k - P_{k - 1}\) | \(=\) | \(\displaystyle \left({- \frac {x^3} {k^2 \pi^2} }\right) \prod_{n \mathop = 1}^{k - 1} \left({1 - \frac {x^2} {n^2 \pi^2} }\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle - \frac {x^3} {k^2 \pi ^2} + O \left({x^5}\right)\) | Big-O Notation |

By Telescoping Series we find that the coefficient of $x^3$ in $P_k$ is given by:

- $(1): \quad \displaystyle - \frac 1 {\pi^2} \sum_{i \mathop = 1}^k \frac 1 {i^2}$

From Euler Formula for Sine Function:

- $\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$

So by taking the limit as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the Power Series Expansion for Sine Function, we can deduce:

- $\displaystyle - \frac 1 {\pi^2} \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = - \frac 1 {3!}$

hence:

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

$\blacksquare$

## 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.*

## Sources

- 1990: William Dunham:
*Journey Through Genius* - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.14$: Euler's Discovery of the Formula $\displaystyle \sum_1^\infty \frac 1 {n^2} = \frac {\pi^2} 6$