Basel Problem/Proof 2

Let
 * $\displaystyle f(x) = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots$.

We note that
 * $\displaystyle f(0)=1-\frac{0}{3!}+\frac{0}{5!}-\frac{0}{7!}+\frac{0}{9!}-\cdots = 1$.

Now for all $x\neq 0$ $\frac{x}{x}=1$. Therefore,
 * $\displaystyle f(x) = \frac{x}{x}f(x)=\frac{x-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots}{x}$.

From the Taylor Series expansion of sine we know that
 * $\displaystyle \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$.

Therefore, for all $x\neq 0$
 * $\displaystyle f(x)=\frac{\sin x}{x}$.

From Euler's formula for the sine function we know that
 * $\displaystyle f(x) = \left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right)  \left( 1 - \frac{x}{2\pi} \right)  \left( 1 + \frac{x}{2\pi} \right)  \left( 1 - \frac{x}{3\pi} \right)  \left( 1 + \frac{x}{3\pi} \right) \cdots = \left[ 1 - \frac{x^2}{\pi^2} \right]  \left[ 1 - \frac{x^2}{4\pi^2} \right]  \left[ 1 - \frac{x^2}{9\pi^2} \right]  \left[ 1 - \frac{x^2}{16\pi^2} \right]\cdots$.

Expanding this infinite product we get
 * $\displaystyle \left[ 1 - \frac{x^2}{\pi^2} \right] \left[ 1 - \frac{x^2}{4\pi^2} \right]  \left[ 1 - \frac{x^2}{9\pi^2} \right]  \left[ 1 - \frac{x^2}{16\pi^2} \right]\cdots = 1 - \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 + \left(\dots\right)x^4-\cdots $.

From this
 * $\displaystyle 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-\cdots = 1 - \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 + \left(\dots\right)x^4-\cdots$.

Now the first element in each infinite sum is equal to the first element in the second and so on. From this we know that
 * $\displaystyle \frac{x^2}{3!} = \left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \frac{1}{16\pi^2} + \cdots \right)x^2 $.

Dividing each side of this equation by $x^2$ and factoring out $\frac{1}{\pi^2}$ on the right side of the equation we obtain
 * $\displaystyle \frac{1}{6} = \frac{1}{\pi^2}\left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \right)$.

Multiplying each side of this equation by $\pi^2$ we obtain
 * $\displaystyle \frac{\pi^2}{6} = \left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots \right)$.

Therefore,
 * $\displaystyle \zeta \left({2}\right) = \sum\limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$.

This proof is based on one from Journey Through Genius.