Basel Problem/Proof 3

Proof
Let $x \in \openint 0 {\dfrac \pi 2}$ and let $n$ be a non-negative integer.

Let $x_k = \dfrac {k \pi} {2 n + 1}$ for $k = 1, 2, \ldots, n$.

Then:
 * $\sin \paren {2 n + 1} x_k = 0$

So we have:


 * $\displaystyle \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} \paren {-1}^r \cot^{2 \paren {n - r} } x_k = 0$

for $k = 1, 2, \ldots, n$.

The numbers $x_k$ are all distinct and in the interval $\openint 0 {\dfrac \pi 2}$.

By Shape of Cotangent Function, $\cot x$ is positive and injective in the interval $\openint 0 {\dfrac \pi 2}$.

Therefore $\cot^2 x$ is also injective in this interval.

Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$.

These numbers are the $n$ distinct roots of the $n$th degree polynomial:


 * $\displaystyle \map f c := \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} (-1)^r c^{n - r}$

By Viète's Formulas, we can calculate the sum of the roots:


 * $\displaystyle \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2 n + 1} 3} {\binom {2 n + 1} 1} = \frac {2 n \paren {2 n - 1} } 6$

From the Difference of Squares of Cosecant and Cotangent $\cot^2 x = \csc^2 x - 1$ we can similarly deduce:


 * $\displaystyle \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2 n \paren {2 n - 1} } 6 + n = \frac {2 n \paren {2 n + 2} } 6$

By Shape of Sine Function, $\sin x$ is positive in the interval $\openint 0 {\dfrac \pi 2}$.

So from the Tangent Inequality $\sin x < x < \tan x$ for $x \in \openint 0 {\dfrac \pi 2}$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval.

Summing this inequality from $x_1$ to $x_n$ gives:

or equivalently:

By Combination Theorem for Limits of Functions, the and  both tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity.

Therefore by the Squeeze Theorem:


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