Editing Basel problem (section)

==The Riemann zeta function ==
The [[Riemann zeta function]] {{math|''ζ''(''s'')}} is one of the most significant functions in mathematics because of its relationship to the distribution of the [[prime number]]s. The zeta function is defined for any [[complex number]] {{math|''s''}} with real part greater than 1 by the following formula:
<math display=block>\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}.</math>

{{anchor|zeta_2}}Taking {{math|''s'' {{=}} 2}}, we see that {{math|''ζ''(2)}} is equal to the sum of the reciprocals of the squares of all positive integers:
<math display=block>\zeta(2) = \sum_{n=1}^\infty \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.</math>

Convergence can be proven by the [[integral test for convergence#Applications|integral test]], or by the following inequality:
<math display=block>\begin{align}
  \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.
\end{align}</math>

This gives us the [[upper bound]] 2, and because the infinite sum contains no negative terms, it must converge to a value strictly between 0 and 2. It can be shown that {{math|''ζ''(''s'')}} has a simple expression in terms of the [[Bernoulli number]]s whenever {{math|''s''}} is a positive even integer. With {{math|''s'' {{=}} 2''n''}}:<ref>{{citation|first1=Tsuneo|last1=Arakawa|first2=Tomoyoshi|last2=Ibukiyama|first3=Masanobu|last3=Kaneko|title=Bernoulli Numbers and Zeta Functions|publisher=Springer|date=2014|page=61|isbn=978-4-431-54919-2}}</ref>
<math display=block>\zeta(2n) = \frac{(2\pi)^{2n}(-1)^{n+1}B_{2n}}{2\cdot(2n)!}.</math>