Editing Basel problem (section)

==A proof using Euler's formula and L'Hôpital's rule==

The normalized [[sinc function]] <math>\text{sinc}(x)=\frac{\sin (\pi x)}{\pi x}</math> has a [[Weierstrass factorization theorem|Weierstrass factorization]] representation as an infinite product:
<math display=block>\frac{\sin (\pi x)}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right).</math>

The infinite product is [[Analytic function|analytic]], so taking the [[natural logarithm]] of both sides and differentiating yields
<math display=block>\frac{\pi \cos (\pi x)}{\sin (\pi x)}-\frac{1}{x}=-\sum_{n=1}^\infty \frac{2x}{n^2-x^2}</math>

(by [[Uniform convergence#To differentiability|uniform convergence]], the interchange of the derivative and infinite series is permissible). After dividing the equation by <math>2x</math> and regrouping one gets
<math display=block>\frac{1}{2x^2}-\frac{\pi \cot (\pi x)}{2x}=\sum_{n=1}^\infty \frac{1}{n^2-x^2}.</math>

We make a change of variables (<math>x=-it</math>):
<math display=block>-\frac{1}{2t^2}+\frac{\pi \cot (-\pi it)}{2it}=\sum_{n=1}^\infty \frac{1}{n^2+t^2}.</math>

[[Euler's formula]] can be used to deduce that
<math display=block>\frac{\pi \cot (-\pi i t)}{2it}=\frac{\pi}{2it}\frac{i\left(e^{2\pi t}+1\right)}{e^{2\pi t}-1}=\frac{\pi}{2t}+\frac{\pi}{t\left(e^{2\pi t} - 1\right)}.</math>
or using the corresponding [[hyperbolic function]]:
<math display=block>\frac{\pi \cot (-\pi i t)}{2it}=\frac{\pi}{2t}{i\cot (\pi i t)}=\frac{\pi}{2t}\coth(\pi t).</math>

Then
<math display=block>\sum_{n=1}^\infty \frac{1}{n^2+t^2}=\frac{\pi \left(te^{2\pi t}+t\right)-e^{2\pi t}+1}{2\left(t^2 e^{2\pi t}-t^2\right)}=-\frac{1}{2t^2} + \frac{\pi}{2t} \coth(\pi t).</math>

Now we take the [[limit (mathematics)|limit]] as <math>t</math> approaches zero and use [[L'Hôpital's rule]] thrice. By [[Tannery's theorem]] applied to <math display="inline">\lim_{t\to\infty}\sum_{n=1}^\infty 1/(n^2+1/t^2)</math>, we can [[Interchange of limiting operations|interchange the limit and infinite series]] so that <math display="inline">\lim_{t\to 0}\sum_{n=1}^\infty 1/(n^2+t^2)=\sum_{n=1}^\infty 1/n^2</math> and by L'Hôpital's rule
<math display=block>\begin{align}\sum_{n=1}^\infty \frac{1}{n^2}&=\lim_{t\to 0}\frac{\pi}{4}\frac{2\pi te^{2\pi t}-e^{2\pi t}+1}{\pi t^2 e^{2\pi t} + te^{2\pi t}-t}\\[6pt]
&=\lim_{t\to 0}\frac{\pi^3 te^{2\pi t}}{2\pi \left(\pi t^2 e^{2\pi t}+2te^{2\pi t} \right)+e^{2\pi t}-1}\\[6pt]
&=\lim_{t\to 0}\frac{\pi^2 (2\pi t+1)}{4\pi^2 t^2+12\pi t+6}\\[6pt]
&=\frac{\pi^2}{6}.\end{align}</math>