Basel Problem/Proof 10

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:

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

$\blacksquare$