Theorem: Basel Problem

:$\ds \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$
where $\zeta$ denotes the Riemann zeta function.

Proof:
For $z \ne 0$, from Mittag-Leffler Expansion for Hyperbolic Cotangent Function, we have: 
:$\ds \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$
Consider the set:
:$A = \set {z \in \C : \size z \le \dfrac 1 2}$. 
Then for each $n \in \N$ and $z \in A$, we have: 
{{begin-eqn}}
{{eqn | l = \size {\frac 1 {z^2 + n^2} }
      | o = \le
      | r = \frac 1 {\size {\size z^2 - n^2} }
      | c = Reverse Triangle Inequality
}}
{{eqn | o = \le
      | r = \frac 1 {n^2 - \frac 1 4}
}}
{{end-eqn}}
For all $n \in \N$, we have: 
:$n^2 - \dfrac 1 4 \ge \dfrac 1 2 n^2$ 
So that: 
:$\dfrac 1 {n^2 - \dfrac 1 4} \le \dfrac 2 {n^2}$ 
We have: 
{{begin-eqn}}
{{eqn | l = \int_1^\infty \frac 2 {x^2} \rd x
      | r = 2 \intlimits {-\frac 1 x} 1 \infty
      | c = Fundamental Theorem of Calculus, Primitive of Power
}}
{{eqn | r = 2
}}
{{eqn | o = <
      | r = \infty
}}
{{end-eqn}}
So, from the Cauchy Integral Test, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 2 {n^2}$ converges.
So, from the {{Corollary|Comparison Test|1}}, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 - \frac 1 4}$ converges.
So by the Weierstrass M-Test, the series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$
converges uniformly on $A$. 
Then from Uniformly Convergent Series of Continuous Functions is Continuous, the function $f : A \to \C$ defined by: 
:$\ds \map f z = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$
is continuous.
So:
:$\ds \lim_{z \mathop \to 0} \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2} = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}$
We can write: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z}
      | r = \frac 1 {2 z} \paren {\frac {\pi \paren {e^{\pi z} + e^{-\pi z} } } {e^{\pi z} - e^{-\pi z} } - \frac 1 z}
      | c = {{Defof|Hyperbolic Cotangent}}
}}
{{eqn | r = \frac 1 {2 z} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {z \paren {e^{2 \pi z} - 1} } }
}}
{{end-eqn}}
so that: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$
We have at $z = 0$: 
{{begin-eqn}}
{{eqn | l = \pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1
      | r = \pi \times 0 \times \paren {e^0 + 1} - e^0 + 1
}}
{{eqn | r = 1 - e^0
}}
{{eqn | r = 0
      | c = Exponential of Zero
}}
{{end-eqn}}
and: 
:$2 z^2 \paren {e^{2 \pi z} + 1} = 0$
So by L'Hopital's Rule:
{{begin-eqn}}
{{eqn | l = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }
      | r = \lim_{z \mathop \to 0} \paren {\frac {\pi \paren {e^{2 \pi z} + 1} + 2 \pi^2 z e^{2 \pi z} - 2 \pi e^{2 \pi z} } {4 z \paren {e^{2 \pi z} - 1} + 4 \pi z^2 e^{2 \pi z} } }
      | c = Product Rule for Derivatives, Derivative of Exponential Function, Derivative of Power
}}
{{eqn | r = \lim_{z \mathop \to 0} \paren {\frac {2 \pi^2 e^{2 \pi z} + 2 \pi^2 e^{2 \pi z} + 4 \pi^3 z e^{2 \pi z} - 4 \pi^2 e^{2 \pi z} } {4 \paren {e^{2 \pi z} - 1} + 8 \pi z e^{2 \pi z} + 8 \pi z e^{2 \pi z} + 8 \pi^2 z^2 e^{2 \pi z} } }
      | c = L'Hopital's Rule
}}
{{eqn | r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {z e^{2 \pi z} } {-4 + e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} } }
}}
{{eqn | r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {e^{2 \pi z} + 2 \pi z e^{2 \pi z} } {2 \pi e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} + e^{2 \pi z} \paren {16 \pi z + 16 \pi} } }
      | c = L'Hopital's Rule
}}
{{eqn | r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {1 + 2 \pi z} {16 \pi^3 z^2 + 32 \pi^2 z + 8 \pi + 16 \pi z + 16 \pi} }
}}
{{eqn | r = \frac {4 \pi^3} {24 \pi}
}}
{{eqn | r = \frac {\pi^2} 6
}}
{{end-eqn}}
giving: 
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
{{qed}}