Basel Problem - The Riemann Zeta Function

The Riemann Zeta Function

The Riemann zeta function is one of the most important functions in mathematics, because of its relationship to the distribution of the prime numbers. The function is defined for any complex number s with real part > 1 by the following formula:

$\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}.$

Taking s = 2, we see that is equal to the sum of the reciprocals of the squares of the positive integers:

$\zeta(2) = \sum_{n=1}^\infin \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6} \approx 1.644934.$

Convergence can be proven with the following inequality:

$\sum_{n=1}^N \frac{1}{n^2} < 1 + \sum_{n=2}^N \frac{1}{n(n-1)} = 1 + \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) = 1 + 1 - \frac{1}{N} \; \stackrel{N \to \infty}{\longrightarrow} \; 2.$

This gives us the upper bound, and because the infinite sum has only positive terms, it must converge. It can be shown that has a nice expression in terms of the Bernoulli numbers whenever s is a positive even integer. With :

$\zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n+1}B_{2n}}{2\cdot(2n)!}$

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