Editing Basel problem (section)

==Euler's approach==
Euler's original derivation of the value <math display="inline">\frac{\pi^2}{6}</math> essentially extended observations about finite [[polynomial]]s and assumed that these same properties hold true for infinite series.

Of course, Euler's original reasoning requires justification (100 years later, [[Karl Weierstrass]] proved that Euler's representation of the sine function as an infinite product is valid, by the [[Weierstrass factorization theorem]]), but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler's argument, recall the [[Taylor series]] expansion of the [[trigonometric function|sine function]]
<math display=block> \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots </math>
Dividing through by {{Mvar|x}} gives
<math display=block> \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots .</math>

The [[Weierstrass factorization theorem]] shows that the right-hand side is the product of linear factors given by its roots, just as for finite polynomials. Euler assumed this as a [[heuristic]] for expanding an infinite degree [[polynomial]] in terms of its roots, but in fact it is not always true for general <math>P(x)</math>.<ref>A priori, since the left-hand-side is a [[polynomial]] (of infinite degree) we can write it as a product of its roots as
<math display=block>\begin{align}
\sin(x) & = x (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2) \cdots \\
     & = Ax \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) \cdots.
\end{align}
</math> Then since we know from elementary [[calculus]] that <math>\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1</math>, we conclude that the leading constant must satisfy <math>A = 1</math>.</ref> This factorization expands the equation into:
<math display=block>\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) \cdots \\
                    &= \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) \cdots
\end{align}</math>

If we formally multiply out this product and collect all the {{math|''x''<sup>2</sup>}} terms (we are allowed to do so because of [[Newton's identities]]), we see by induction that the {{math|''x''<sup>2</sup>}} coefficient of {{math|{{sfrac|sin ''x''|''x''}}}} is <ref>In particular, letting <math>H_n^{(2)} := \sum_{k=1}^n k^{-2}</math> denote a [[generalized harmonic number|generalized second-order harmonic number]], we can easily prove by [[Mathematical induction|induction]] that <math>[x^2] \prod_{k=1}^{n} \left(1-\frac{x^2}{\pi^2}\right) = -\frac{H_n^{(2)}}{\pi^2} \rightarrow -\frac{\zeta(2)}{\pi^2}</math> as <math>n \rightarrow \infty</math>.</ref>
<math display=block> -\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.</math>

But from the original infinite series expansion of {{math|{{sfrac|sin ''x''|''x''}}}}, the coefficient of {{math|''x''<sup>2</sup>}} is {{math|−{{sfrac|1|3!}} {{=}} −{{sfrac|1|6}}}}. These two coefficients must be equal; thus,
<math display=block>-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.</math>

Multiplying both sides of this equation by −{{pi}}<sup>2</sup> gives the sum of the reciprocals of the positive square integers.<ref name="HAVIL-GAMMA">{{citation|last1=Havil|first1=J.|title=Gamma: Exploring Euler's Constant|url=https://archive.org/details/gammaexploringeu00havi_882|url-access=limited|date=2003|publisher=Princeton University Press|location=Princeton, New Jersey|isbn=0-691-09983-9|pages=[https://archive.org/details/gammaexploringeu00havi_882/page/n60 37]–42 (Chapter 4)}}</ref>
<math display=block>\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.</math>
===Generalizations of Euler's method using elementary symmetric polynomials===

Using formulae obtained from [[elementary symmetric polynomial]]s,<ref>Cf., the formulae for generalized Stirling numbers proved in: {{citation|last1=Schmidt|first1=M. D.|title=Combinatorial Identities for Generalized Stirling Numbers Expanding f-Factorial Functions and the f-Harmonic Numbers|journal=J. Integer Seq.|date=2018|volume=21|issue=Article 18.2.7|url=https://cs.uwaterloo.ca/journals/JIS/VOL21/Schmidt/schmidt18.html}}</ref> this same approach can be used to enumerate formulae for the even-indexed [[zeta constants|even zeta constants]] which have the following known formula expanded by the [[Bernoulli numbers]]:
<math display=block>\zeta(2n) = \frac{(-1)^{n-1} (2\pi)^{2n}}{2 \cdot (2n)!} B_{2n}. </math>

For example, let the partial product for <math>\sin(x)</math> expanded as above be defined by <math>\frac{S_n(x)}{x} := \prod\limits_{k=1}^n \left(1 - \frac{x^2}{k^2 \cdot \pi^2}\right)</math>. Then using known [[Newton's identities#Expressing elementary symmetric polynomials in terms of power sums|formulas for elementary symmetric polynomial]]s (a.k.a., Newton's formulas expanded in terms of [[power sum]] identities), we can see (for example) that
<math display=block>
\begin{align}
\left[x^4\right] \frac{S_n(x)}{x} & = \frac{1}{2\pi^4}\left(\left(H_n^{(2)}\right)^2 - H_n^{(4)}\right) \qquad \xrightarrow{n \rightarrow \infty} \qquad \frac{1}{2\pi^4}\left(\zeta(2)^2-\zeta(4)\right) \\[4pt]
& \qquad \implies \zeta(4) = \frac{\pi^4}{90} = -2\pi^4 \cdot [x^4] \frac{\sin(x)}{x} +\frac{\pi^4}{36} \\[8pt]
\left[x^6\right] \frac{S_n(x)}{x} & = -\frac{1}{6\pi^6}\left(\left(H_n^{(2)}\right)^3 - 2H_n^{(2)} H_n^{(4)} + 2H_n^{(6)}\right) \qquad \xrightarrow{n \rightarrow \infty} \qquad \frac{1}{6\pi^6}\left(\zeta(2)^3-3\zeta(2)\zeta(4) + 2\zeta(6)\right) \\[4pt]
& \qquad \implies \zeta(6) = \frac{\pi^6}{945} = -3 \cdot \pi^6 [x^6] \frac{\sin(x)}{x} - \frac{2}{3} \frac{\pi^2}{6} \frac{\pi^4}{90} + \frac{\pi^6}{216},
\end{align}
</math>

and so on for subsequent coefficients of <math>[x^{2k}] \frac{S_n(x)}{x}</math>. There are [[Newton's identities#Expressing power sums in terms of elementary symmetric polynomials|other forms of Newton's identities]] expressing the (finite) power sums <math>H_n^{(2k)}</math> in terms of the [[elementary symmetric polynomial]]s, <math>e_i \equiv e_i\left(-\frac{\pi^2}{1^2}, -\frac{\pi^2}{2^2}, -\frac{\pi^2}{3^2}, -\frac{\pi^2}{4^2}, \ldots\right), </math> but we can go a more direct route to expressing non-recursive formulas for <math>\zeta(2k)</math> using the method of [[elementary symmetric polynomial]]s. Namely, we have a recurrence relation between the elementary symmetric polynomials and the [[Power sum symmetric polynomial|power sum polynomials]] given as on [[Newton's identities#Comparing coefficients in series|this page]] by
<math display=block>(-1)^{k}k e_k(x_1,\ldots,x_n) = \sum_{j=1}^k (-1)^{k-j-1} p_j(x_1,\ldots,x_n)e_{k-j}(x_1,\ldots,x_n),</math>

which in our situation equates to the limiting recurrence relation (or [[generating function]] convolution, or [[Cauchy product|product]]) expanded as
<math display=block> \frac{\pi^{2k}}{2}\cdot \frac{(2k) \cdot (-1)^k}{(2k+1)!} = -[x^{2k}] \frac{\sin(\pi x)}{\pi x} \times \sum_{i \geq 1} \zeta(2i) x^i. </math>

Then by differentiation and rearrangement of the terms in the previous equation, we obtain that
<math display=block>\zeta(2k) = [x^{2k}]\frac{1}{2}\left(1-\pi x\cot(\pi x)\right). </math>

===Consequences of Euler's proof===
By the above results, we can conclude that <math>\zeta(2k)</math> is ''always'' a [[rational]] multiple of <math>\pi^{2k}</math>. In particular, since <math>\pi</math> and integer powers of it are [[Transcendental number|transcendental]], we can conclude at this point that <math>\zeta(2k)</math> is [[irrational]], and more precisely, [[Transcendental number|transcendental]] for all <math>k \geq 1</math>. By contrast, the properties of the odd-indexed [[zeta constants]], including [[Apéry's constant]] <math>\zeta(3)</math>, are almost completely unknown.