Basel Problem/Proof 10

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Theorem

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

where $\zeta$ denotes the Riemann zeta function.


Proof

From Mittag-Leffler Expansion for Hyperbolic Cotangent Function, we have:

$\displaystyle \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$

We can write:

\(\displaystyle \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z}\) \(=\) \(\displaystyle \frac 1 {2 z} \paren {\frac {\pi \paren {e^{\pi z} + e^{-\pi z} } } {e^{\pi z} - e^{-\pi z} } - \frac 1 z}\) Definition of Hyperbolic Cotangent
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 z} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {z \paren {e^{2 \pi z} - 1} } }\)

We therefore have:

$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$

We have at $z = 0$:

\(\displaystyle \pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1\) \(=\) \(\displaystyle \pi \times 0 \times \paren {e^0 + 1} - e^0 + 1\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - e^0\)
\(\displaystyle \) \(=\) \(\displaystyle 0\) Exponential of Zero

and:

$2 z^2 \paren {e^{2 \pi z} + 1} = 0$

So by L'Hopital's Rule:

\(\displaystyle \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }\) \(=\) \(\displaystyle \lim_{z \mathop \to 0} \paren {\frac {\pi \paren {e^{2 \pi z} + 1} + 2 \pi^2 z e^{2 \pi z} - 2 \pi e^{2 \pi z} } {4 z \paren {e^{2 \pi z} - 1} + 4 \pi z^2 e^{2 \pi z} } }\) Product Rule, Derivative of Exponential Function, Derivative of Power
\(\displaystyle \) \(=\) \(\displaystyle \lim_{z \mathop \to 0} \paren {\frac {2 \pi^2 e^{2 \pi z} + 2 \pi^2 e^{2 \pi z} + 4 \pi^3 z e^{2 \pi z} - 4 \pi^2 e^{2 \pi z} } {4 \paren {e^{2 \pi z} - 1} + 8 \pi z e^{2 \pi z} + 8 \pi z e^{2 \pi z} + 8 \pi^2 z^2 e^{2 \pi z} } }\) L'Hopital's Rule
\(\displaystyle \) \(=\) \(\displaystyle 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {z e^{2 \pi z} } {-4 + e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {e^{2 \pi z} + 2 \pi z e^{2 \pi z} } {2 \pi e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} + e^{2 \pi z} \paren {16 \pi z + 16 \pi} } }\) L'Hopital's Rule
\(\displaystyle \) \(=\) \(\displaystyle 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {1 + 2 \pi z} {16 \pi^3 z^2 + 32 \pi^2 z + 8 \pi + 16 \pi z + 16 \pi} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \pi^3} {24 \pi}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 6\)

giving:

$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$

$\blacksquare$