Basel Problem/Proof 5

Proof
Let the function $f_n$ be defined as:


 * $(1): \quad \map {f_n} x := \dfrac 1 2 + \cos x + \cos 2 x + \cdots + \cos n x$

By Sum of Cosines of Multiples of Angle:
 * $(2): \quad \map {f_n} x = \dfrac {\map \sin {\paren {2 n + 1} x / 2} } {2 \map \sin {x / 2} }$

Let $E_n$ be defined as:

The terms for even $k$ on the {{{RHS}} are zero, so:
 * $\ds (4): \quad \dfrac 1 2 E_{2 n - 1} = \dfrac {\pi^2} 8 - \sum_{k \mathop = 1}^n \dfrac 1 {\paren {2 k - 1}^2}$

It remains to be shown that:
 * $\ds \lim_{n \mathop \to \infty} E_{2 n - 1} = 0$

which will establish:
 * $\ds (5): \quad \dfrac {\pi^2} 8 = \sum_{k \mathop = 1}^n \dfrac 1 {\paren {2 k - 1}^2}$

Using $(2)$, let $\map g x$ be the function defined as:
 * $\map g x := \map {\dfrac \d {\d x} } {\dfrac {x / 2} {\map \sin {x /2} } }$

Using Integration by Parts, we obtain:
 * $\ds (6): \quad E_{2 n - 1} = \dfrac 1 {4 n - 1} \paren {2 + 2 \int_0^\pi \map g x \cos \dfrac {\paren {4 n - 1} x} 2 \rd x}$

during which we use Limit of $\dfrac {\sin x} x$: Corollary:
 * $\ds \lim_{x \mathop \to 0} \dfrac {x / 2} {\map \sin {x / 2} } = 1$

We have that $\map g x$ is increasing on the interval of integration.

Therefore $\map g x$ is bounded on the interval $\closedint 0 \pi$ by $\map g \pi = \dfrac \pi 2$.

Hence $(5)$ has been established as being true.

Now we divide the (strictly) positive integers into even and odd, and use $(5)$ to obtain: