Basel Problem/Proof 10
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Theorem
- $\ds \map \zeta 2 = \sum_{n \mathop = 1}^\infty {\frac 1 {n^2} } = \frac {\pi^2} 6$
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:
\(\ds \size {\frac 1 {z^2 + n^2} }\) | \(\le\) | \(\ds \frac 1 {\size {\size z^2 - n^2} }\) | Reverse Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {n^2 - \frac 1 4}\) |
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:
\(\ds \int_1^\infty \frac 2 {x^2} \rd x\) | \(=\) | \(\ds 2 \intlimits {-\frac 1 x} 1 \infty\) | Fundamental Theorem of Calculus, Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \infty\) |
So, from the Cauchy Integral Test, we have:
- $\ds \sum_{n \mathop = 1}^\infty \frac 2 {n^2}$ converges.
So, from the Comparison Test: Corollary, 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:
\(\ds \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z}\) | \(=\) | \(\ds \frac 1 {2 z} \paren {\frac {\pi \paren {e^{\pi z} + e^{-\pi z} } } {e^{\pi z} - e^{-\pi z} } - \frac 1 z}\) | Definition of Hyperbolic Cotangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 z} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {z \paren {e^{2 \pi z} - 1} } }\) |
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 $z = 0$:
\(\ds \pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1\) | \(=\) | \(\ds \pi \times 0 \times \paren {e^0 + 1} - e^0 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - e^0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Exponential of Zero |
and:
- $2 z^2 \paren {e^{2 \pi z} + 1} = 0$
So by L'Hopital's Rule:
\(\ds \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }\) | \(=\) | \(\ds \lim_{z \mathop \to 0} \paren {\frac {\pi \paren {e^{2 \pi z} + 1} + 2 \pi^2 z e^{2 \pi z} - 2 \pi e^{2 \pi z} } {4 z \paren {e^{2 \pi z} - 1} + 4 \pi z^2 e^{2 \pi z} } }\) | Product Rule for Derivatives, Derivative of Exponential Function, Derivative of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to 0} \paren {\frac {2 \pi^2 e^{2 \pi z} + 2 \pi^2 e^{2 \pi z} + 4 \pi^3 z e^{2 \pi z} - 4 \pi^2 e^{2 \pi z} } {4 \paren {e^{2 \pi z} - 1} + 8 \pi z e^{2 \pi z} + 8 \pi z e^{2 \pi z} + 8 \pi^2 z^2 e^{2 \pi z} } }\) | L'Hopital's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {z e^{2 \pi z} } {-4 + e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {e^{2 \pi z} + 2 \pi z e^{2 \pi z} } {2 \pi e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} + e^{2 \pi z} \paren {16 \pi z + 16 \pi} } }\) | L'Hopital's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {1 + 2 \pi z} {16 \pi^3 z^2 + 32 \pi^2 z + 8 \pi + 16 \pi z + 16 \pi} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \pi^3} {24 \pi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\pi^2} 6\) |
giving:
- $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$
$\blacksquare$