Basel Problem/Proof 5

Proof
Let the function $f_n$ be defined as:


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

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

Let $E_n$ be defined as:

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

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

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

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

Using Integration by Parts, we obtain:
 * $\displaystyle (6): \quad E_{2 n - 1} = \dfrac 1 {4 n - 1} \left({2 + 2 \int_0^\pi g \left({x}\right) \cos \dfrac {\left({4 n - 1}\right) x} 2 \, \mathrm d x}\right)$

during which we use Limit of $\dfrac {\sin x} x$: Corollary:
 * $\displaystyle \lim_{x \mathop \to 0} \dfrac {x / 2} {\sin \left({x / 2}\right)} = 1$

We have that $g \left({x}\right)$ is increasing on the interval of integration.

Therefore $g \left({x}\right)$ is bounded on the interval $\left[{ 0 \,.\,.\, \pi}\right]$ by $g \left({\pi}\right) = \dfrac \pi 2$.

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

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