Editing Basel problem (section)

===The proof===
[[File:limit circle FbN.jpeg|thumb|The inequality<br>
<math>\tfrac{1}{2}r^2\tan\theta > \tfrac{1}{2}r^2\theta > \tfrac{1}{2}r^2\sin\theta</math><br>
is shown pictorially for any <math>\theta \in (0, \pi/2)</math>. The three terms are the areas of the triangle OAC, circle section OAB, and the triangle OAB.

Taking reciprocals and squaring gives<br>
<math>\cot^2\theta<\tfrac{1}{\theta^2}<\csc^2\theta</math>.]]
The main idea behind the proof is to bound the partial (finite) sums
<math display=block>\sum_{k=1}^m \frac{1}{k^2} = \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}</math>
between two expressions, each of which will tend to {{sfrac|{{pi}}<sup>2</sup>|6}} as {{math|''m''}} approaches infinity. The two expressions are derived from identities involving the [[cotangent]] and [[cosecant]] functions. These identities are in turn derived from [[de Moivre's formula]], and we now turn to establishing these identities.

Let {{math|''x''}} be a real number with {{math|0 < ''x'' < {{sfrac|{{pi}}|2}}}}, and let {{math|''n''}} be a positive odd integer. Then from de Moivre's formula and the definition of the cotangent function, we have
<math display=block>\begin{align}
  \frac{\cos (nx) + i \sin (nx)}{\sin^n x} &= \frac{(\cos x + i\sin x)^n}{\sin^n x} \\[4pt]
                                             &= \left(\frac{\cos x + i \sin x}{\sin x}\right)^n \\[4pt]
                                             &= (\cot x + i)^n.
\end{align}</math>

From the [[binomial theorem]], we have
<math display=block>\begin{align}
(\cot x + i)^n
= & {n \choose 0} \cot^n x + {n \choose 1} (\cot^{n - 1} x)i + \cdots + {n \choose {n - 1}} (\cot x)i^{n - 1} + {n \choose n} i^n \\[6pt]
= & \Bigg( {n \choose 0} \cot^n x - {n \choose 2} \cot^{n - 2} x \pm \cdots \Bigg) \; + \; i\Bigg( {n \choose 1} \cot^{n-1} x - {n \choose 3} \cot^{n - 3} x \pm \cdots \Bigg).
\end{align}</math>

Combining the two equations and equating imaginary parts gives the identity
<math display=block>\frac{\sin (nx)}{\sin^n x} = \Bigg( {n \choose 1} \cot^{n - 1} x - {n \choose 3} \cot^{n - 3} x \pm \cdots \Bigg).</math>

We take this identity, fix a positive integer {{math|''m''}}, set {{math|''n'' {{=}} 2''m'' + 1}}, and consider {{math|''x<sub>r</sub>'' {{=}} {{sfrac|''r''{{pi}}|2''m'' + 1}}}} for {{math|''r'' {{=}} 1, 2, ..., ''m''}}. Then {{math|''nx<sub>r</sub>''}} is a multiple of {{pi}} and therefore {{math|sin(''nx<sub>r</sub>'') {{=}} 0}}. So,
<math display=block>0 = {{2m + 1} \choose 1} \cot^{2m} x_r - {{2m + 1} \choose 3} \cot^{2m - 2} x_r \pm \cdots + (-1)^m{{2m + 1} \choose {2m + 1}}</math>

for every {{math|''r'' {{=}} 1, 2, ..., ''m''}}. The values {{math|''x<sub>r</sub>'' {{=}} ''x''<sub>1</sub>, ''x''<sub>2</sub>, ..., ''x<sub>m</sub>''}} are distinct numbers in the interval {{math|0 < {{math|''x<sub>r</sub>''}} < {{sfrac|{{pi}}|2}}}}. Since the function {{math|cot<sup>2</sup> ''x''}} is [[Injective function|one-to-one]] on this interval, the numbers {{math|''t<sub>r</sub>'' {{=}} cot<sup>2</sup> ''x<sub>r</sub>''}} are distinct for {{math|''r'' {{=}} 1, 2, ..., ''m''}}. By the above equation, these {{math|''m''}} numbers are the roots of the {{math|''m''}}th degree polynomial
<math display=block>p(t) = {{2m + 1} \choose 1}t^m - {{2m + 1} \choose 3}t^{m - 1} \pm \cdots + (-1)^m{{2m+1} \choose {2m + 1}}.</math>

By [[Vieta's formulas]] we can calculate the sum of the roots directly by examining the first two coefficients of the polynomial, and this comparison shows that
<math display=block>\cot ^2 x_1 + \cot ^2 x_2 + \cdots + \cot ^2 x_m = \frac{\binom{2m + 1}3} {\binom{2m + 1}1} = \frac{2m(2m - 1)}6.</math>

Substituting the [[list of trigonometric identities|identity]] {{math|csc<sup>2</sup> ''x'' {{=}} cot<sup>2</sup> ''x'' + 1}}, we have
<math display=block>\csc ^2 x_1 + \csc ^2 x_2 + \cdots + \csc ^2 x_m = \frac{2m(2m - 1)}6 + m = \frac{2m(2m + 2)}6.</math>

Now consider the inequality {{math|cot<sup>2</sup> ''x'' < {{sfrac|1|''x''<sup>2</sup>}} < csc<sup>2</sup> ''x''}} (illustrated geometrically above). If we add up all these inequalities for each of the numbers {{math|''x<sub>r</sub>'' {{=}} {{sfrac|''r''{{pi}}|2''m'' + 1}}}}, and if we use the two identities above, we get
<math display=block>\frac{2m(2m - 1)}6 < \left(\frac{2m + 1}{\pi} \right)^2 + \left(\frac{2m + 1}{2\pi} \right)^2 + \cdots + \left(\frac{2m + 1}{m \pi} \right)^2 < \frac{2m(2m + 2)}6.</math>

Multiplying through by {{math|<big><big>(</big></big>{{sfrac|{{pi}}|2''m'' + 1}}<big><big>)</big></big>{{su|p=2}}}}, this becomes
<math display=block>\frac{\pi ^2}{6}\left(\frac{2m}{2m + 1}\right)\left(\frac{2m - 1}{2m + 1}\right) < \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2} < \frac{\pi ^2}{6}\left(\frac{2m}{2m + 1}\right)\left(\frac{2m + 2}{2m + 1}\right).</math>

As {{math|''m''}} approaches infinity, the left and right hand expressions each approach {{sfrac|{{pi}}<sup>2</sup>|6}}, so by the [[squeeze theorem]],
<math display=block>\zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} =
  \lim_{m \to \infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}\right) = \frac{\pi ^2}{6}</math>

and this completes the proof.

==Proof assuming Weil's conjecture on Tamagawa numbers==
A proof is also possible assuming [[Weil's conjecture on Tamagawa numbers]].<ref>{{citation|title=Algebraic groups and number theory|translator=Rachel Rowen|publisher=Academic Press|author1=[[Vladimir Platonov]]|author2=Andrei Rapinchuk|year=1994}}|</ref>  The conjecture asserts for the case of the [[algebraic group]] SL<sub>2</sub>('''R''') that the [[Tamagawa number]] of the group is one.  That is, the quotient of the special linear group over the rational [[Adele ring|adeles]] by the special linear group of the rationals (a [[compact set]], because <math>SL_2(\mathbb Q)</math> is a lattice in the adeles) has Tamagawa measure 1:
<math display="block">\tau(SL_2(\mathbb Q)\setminus SL_2(A_{\mathbb Q}))=1.</math> 

To determine a Tamagawa measure, the group <math>SL_2</math> consists of matrices
<math display="block">\begin{bmatrix}x&y\\z&t\end{bmatrix}</math>
with <math>xt-yz=1</math>.  An invariant [[volume form]] on the group is
<math display="block">\omega = \frac1x dx\wedge dy\wedge dz.</math>

The measure of the quotient is the product of the measures of <math>SL_2(\mathbb Z)\setminus SL_2(\mathbb R)</math> corresponding to the infinite place, and the measures of <math>SL_2(\mathbb Z_p)</math> in each finite place, where <math>\mathbb Z_p</math> is the [[p-adic integers]].  

For the local factors,
<math display="block">\omega(SL_2(\mathbb Z_p)) = |SL_2(F_p)|\omega(SL_2(\mathbb Z_p,p))</math>
where <math>F_p</math> is the field with <math>p</math> elements, and <math>SL_2(\mathbb Z_p,p)</math> is the [[congruence subgroup]] modulo <math>p</math>.  Since each of the coordinates <math>x,y,z</math> map the latter group onto <math>p\mathbb Z_p</math> and <math>\left|\frac1x\right|_p=1</math>, the measure of <math>SL_2(\mathbb Z_p,p)</math> is <math>\mu_p(p\mathbb Z_p)^3=p^{-3}</math>, where <math>\mu_p</math> is the normalized [[Haar measure]] on <math>\mathbb Z_p</math>.  Also, a standard computation shows that <math>|SL_2(F_p)|=p(p^2-1)</math>.  Putting these together gives <math>\omega(SL_2(\mathbb Z_p))=(1-1/p^2)</math>.

At the infinite place, an integral computation over the fundamental domain of <math>SL_2(\mathbb Z)</math> shows that <math>\omega(SL_2(\mathbb Z)\setminus SL_2(\mathbb R)=\pi^2/6</math>, and therefore the Weil conjecture finally gives
<math display="block">1 = \frac{\pi^2}6\prod_p \left(1-\frac1{p^2}\right).</math>
On the right-hand side, we recognize the [[Euler product]] for <math>1/\zeta(2)</math>, and so this gives the solution to the Basel problem.

This approach shows the connection between (hyperbolic) geometry and arithmetic, and can be inverted to give a proof of the Weil conjecture for the special case of <math>SL_2</math>, contingent on an independent proof that <math>\zeta(2)=\pi^2/6</math>.

==Geometric proof==


The Basel problem can be proved with [[Euclidean geometry]], using the insight that ''the real line can be seen as a circle of infinite radius''. An intuitive, if not completely rigorous, sketch is given here.

* Choose an integer <math>N</math>, and take <math>N</math> equally spaced points on a circle with ''circumference'' equal to <math>2N</math>. The radius of the circle is <math>N/\pi</math> and the length of each [[Circular arc|arc]] between two points is <math>2</math>. Call the points <math>P_{1..N}</math>.
* Take another generic point <math>Q</math> on the circle, which will lie at a fraction <math>0 < \alpha < 1</math> of the arc between two consecutive points (say <math>P_1</math> and <math>P_2</math> without loss of generality).
* Draw all the [[Chord (geometry)|chords]] joining <math>Q</math> with each of the <math>P_{1..N}</math> points. Now (this is the key to the proof), compute the ''sum of the inverse squares'' of the lengths of all these chords, call it <math>sisc</math>.
* The proof relies on the notable fact that (for a fixed <math>\alpha</math>), ''the <math>sisc</math> does not depend on <math>N</math>.'' Note that intuitively, as <math>N</math> increases, the number of chords increases, but their length increases too (as the circle gets bigger), so their inverse square decreases.
* In particular, take the case where <math>\alpha = 1/2</math>, meaning that <math>Q</math> is the midpoint of the arc between two consecutive <math>P</math>'s. The <math>sisc</math> can then be found trivially from the case <math>N=1</math>, where there is only one <math>P</math>, and one <math>Q</math> on the opposite side of the circle. Then the chord is the diameter of the circle, of length <math>2/\pi</math>. The <math>sisc</math> is then <math>\pi^2/4</math>.
* When <math>N</math> goes to infinity, the circle approaches the real line. If you set the origin at <math>Q</math>, the points <math>P_{1..N}</math> are positioned at the ''odd'' integer positions (positive and negative), since the arcs have length 1 from <math>Q</math> to <math>P_1</math>, and 2 onward. You hence get this variation of the Basel Problem:

<math display="block">
\sum_{z=-\infty}^{\infty} \frac{1}{(2z-1)^2} = \frac{\pi^2}{4}
</math>

* From here, you can recover the original formulation with a bit of algebra, as:

<math display="block">
    \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{1}{2}\sum_{z=-\infty}^{\infty} \frac{1}{(2z-1)^2} + \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}
</math>

that is,

<math display="block">
    \frac{3}{4}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{8}
</math>

or 

<math display="block">
    \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
</math>.

The independence of the <math>sisc</math> from <math>N</math> can be proved easily with Euclidean geometry for the more restrictive case where <math>N</math> is a power of 2, i.e. <math>N = 2^n</math>, which still allows the limiting argument to be applied. The proof proceeds by [[Mathematical induction|induction]] on <math>n</math>, and uses the [[Inverse Pythagorean theorem|Inverse Pythagorean Theorem]], which states that:

<math display="block">
\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{h^2}
</math>

where <math>a</math> and <math>b</math> are the legs and <math>h</math> is the height of a right triangle.

* In the base case of <math>n=0</math>, there is only 1 chord. In the case of <math>\alpha = 1/2</math>, it corresponds to the diameter and the <math>sisc</math> is <math>\pi^2/4</math> as stated above.
* Now, assume that you have <math>2^n</math> points on a circle with radius <math>2^n/\pi</math> and center <math>O</math>, and <math>2^{n+1}</math> points on a circle with radius <math>2^{n+1}/\pi</math> and center <math>R</math>. The induction step consists in showing that these 2 circles have the same <math>sisc</math> for a given <math>\alpha</math>.

* Start by drawing the circles so that they share point <math>Q</math>. Note that <math>R</math> lies on the smaller circle. Then, note that <math>2^{n+1}</math> is always even, and a simple geometric argument shows that you can pick ''pairs'' of opposite points <math>P_1</math> and <math>P_2</math> on the larger circle by joining each pair with a diameter. Furthermore, for each pair, one of the points will be in the "lower" half of the circle (closer to <math>Q</math>) and the other in the "upper" half.
[[File:Induction step 1728668638527.jpg|thumb|The sum of inverse squares of distances of P1 and P2 from Q equals the inverse square distance from P to Q.]]
* The diameter of the bigger circle <math>P_1P_2</math> cuts the smaller circle at <math>R</math> and at another point <math>P</math>. You can then make the following considerations:
** <math>P_1 \widehat{Q} P_2</math> is a right angle, since <math>P_1P_2</math> is a diameter.
** <math>Q \widehat{P} R</math> is a right angle, since <math>QR</math> is a diameter.
** <math>Q \widehat{R} P_2 = Q \widehat{R} P</math> is half of <math>Q \widehat{O} P</math> for the [[Inscribed angle|Inscribed Angle Theorem]].
** Hence, the arc <math>QP</math> is equal to the arc <math>QP_2</math>, again because the radius is half.
** The chord <math>QP</math> is the height of the right triangle <math>QP_1P_2</math>, hence for the Inverse Pythagorean Theorem:
<math display="block">
\frac{1}{\overline{QP}^2} = \frac{1}{\overline{QP_1}^2} + \frac{1}{\overline{QP_2}^2}
</math>


* Hence for half of the points on the bigger circle (the ones in the lower half) there is a corresponding point on the smaller circle with the same arc distance from <math>Q</math> (since the circumference of the smaller circle is half that of the bigger circle, the last two points closer to <math>R</math> must have arc distance 2 as well). Vice versa, for each of the <math>2^n</math> points on the smaller circle, we can build a pair of points on the bigger circle, and all of these points are equidistant and have the same arc distance from <math>Q</math>.
* Furthermore, the total <math>sisc</math> for the bigger circle is the same as the <math>sisc</math> for the smaller circle, since each pair of points on the bigger circle has the same inverse square sum as the corresponding point on the smaller circle.<ref>{{cite web |url= https://www.math.chalmers.se/~wastlund/Cosmic.pdf |title= Summing Inverse Squares by Euclidean Geometry |author= Johan Wästlund |date= December 8, 2010 |access-date= 2024-10-11 |website= Chalmers University of Technology |publisher= Department of Mathematics, Chalmers University}}</ref>

==Other identities==
See the special cases of the identities for the [[Riemann zeta function#Representations|Riemann zeta function]] when <math>s = 2.</math> Other notably special identities and representations of this constant appear in the sections below.

===Series representations===
The following are series representations of the constant:<ref name="MWZETA2">{{mathworld|title=Riemann Zeta Function \zeta(2)|id=RiemannZetaFunctionZeta2|mode=cs2}}</ref>
<math display=block>\begin{align}
  \zeta(2) &= 3 \sum_{k=1}^\infty \frac{1}{k^2 \binom{2k}{k}} \\[6pt]
     &= \sum_{i=1}^\infty \sum_{j=1}^\infty \frac{(i-1)! (j-1)!}{(i+j)!}.
 \end{align}</math>

There are also [[Bailey–Borwein–Plouffe formula|BBP-type]] series expansions for {{math|''&zeta;''(2)}}.<ref name="MWZETA2" />

===Integral representations===

The following are integral representations of <math>\zeta(2)\text{:}</math><ref>{{cite arXiv|last1=Connon|first1=D. F.|title=Some series and integrals involving the Riemann zeta function, binomial coefficients and the harmonic numbers (Volume I)|year=2007|class=math.HO|eprint=0710.4022|mode=cs2}}</ref><ref>{{mathworld|title=Double Integral|id=DoubleIntegral|mode=cs2}}</ref><ref>{{mathworld|title=Hadjicostas's Formula|id=HadjicostassFormula.html|mode=cs2}}</ref>
<math display=block> \begin{align}
  \zeta(2) & = -\int_0^1 \frac{\log x}{1-x} \, dx \\[6pt]
     & = \int_0^{\infty} \frac{x}{e^x-1} \, dx \\[6pt]
     & = \int_0^1 \frac{(\log x)^2}{(1+x)^2} \, dx \\[6pt]
     & = 2 + 2\int_1^{\infty} \frac{\lfloor x \rfloor -x}{x^3} \, dx \\[6pt]
     & = \exp\left(2 \int_2^{\infty} \frac{\pi(x)}{x(x^2-1)} \,dx\right) \\[6pt]
     & = \int_0^1 \int_0^1 \frac{dx \, dy}{1-xy} \\[6pt]
     & = \frac{4}{3} \int_0^1 \int_0^1 \frac{dx \, dy}{1-(xy)^2} \\[6pt]
     & = \int_0^1 \int_0^1 \frac{1-x}{1-xy} \, dx \, dy + \frac{2}{3}.
 \end{align}</math>

===Continued fractions===
In van der Poorten's classic article chronicling [[Apéry's constant|Apéry's proof of the irrationality of <math>\zeta(3)</math>]],<ref>{{Citation
 |first=Alfred
 |last=van der Poorten
 |author-link=Alfred van der Poorten
 |title=A proof that Euler missed ... Apéry's proof of the irrationality of {{math|''ζ''(3)}}
 |journal=[[The Mathematical Intelligencer]]
 |volume=1
 |issue=4
 |year=1979
 |pages=195–203
 |doi=10.1007/BF03028234
 |s2cid=121589323
 |url=http://www.maths.mq.edu.au/~alf/45.pdf
 |url-status=dead
 |archive-url=https://web.archive.org/web/20110706114957/http://www.maths.mq.edu.au/~alf/45.pdf
 |archive-date=2011-07-06
}}</ref> the author notes as "a red herring" the similarity of a [[simple continued fraction]] for Apery's constant, and the following one for the Basel constant:
<math display=block>\frac{\zeta(2)}{5} = \cfrac{1}{\widetilde{v}_1 - \cfrac{1^4}{\widetilde{v}_2-\cfrac{2^4}{\widetilde{v}_3-\cfrac{3^4}{\widetilde{v}_4-\ddots}}}}, </math>
where <math>\widetilde{v}_n = 11n^2-11n+3 \mapsto \{3,25,69,135,\ldots\}</math>. Another continued fraction of a similar form is:<ref name="Berndt">{{citation |last1=Berndt |first1=Bruce C. |title=Ramanujan's Notebooks: Part II |date=1989 |publisher=Springer-Verlag |isbn=978-0-387-96794-3 |page=150}}</ref>
<math display=block>\frac{\zeta(2)}{2} = \cfrac{1}{v_1 - \cfrac{1^4}{v_2-\cfrac{2^4}{v_3-\cfrac{3^4}{v_4-\ddots}}}}, </math>
where  <math>v_n = 2n-1 \mapsto \{1,3,5,7,9,\ldots\}</math>.