Question

Show that the series summation of 1/a^2 from a=1 to infinity converge and find the sum of the series. Hence or otherwise prove the summation from a=1 to infinity is equal to π^2/6

Answer 1

The series $\displaystyle\sum_{a=1}^{\infin}\dfrac{1}{a^2}$ converges (p-series with $p=2$ ).

The Taylor series for the function $f(x)=\sin(x)$ is

$$\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$$

Since the equation $f(x)=\sin(x)=0$ has the following roots

$$x=0, x=\pm\pi, x=\pm2\pi,x=\pm3\pi, ...$$

we have

$$\sin(x)=Cx(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)...$$

$$\sin(x)=Cx(x^2-\pi^2)(x^2-2^2\pi^2)(x^2-3^2\pi^2)...$$

Then

$$\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$$

$$=Cx(x^2-\pi^2)(x^2-2^2\pi^2)(x^2-3^2\pi^2)...$$

$$1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+...$$

$$=C(x^2-\pi^2)(x^2-2^2\pi^2)(x^2-3^2\pi^2)...$$

Substitute $x^2$ by $x,$ we have

$$1-\dfrac{x}{3!}+\dfrac{x^2}{5!}-\dfrac{x^3}{7!}+...$$

$$=C(x-\pi^2)(x-2^2\pi^2)(x-3^2\pi^2)...$$

By normalization, we obtain

$$1-\dfrac{x}{3!}+\dfrac{x^2}{5!}-\dfrac{x^3}{7!}+...$$

$$=\dfrac{(x-\pi^2)}{-\pi^2}\cdot\dfrac{(x-2^2\pi^2)}{-2^2\pi^2}\cdot\dfrac{(x-3^2\pi^2)}{-3^2\pi^2}...$$

$$=(1-\dfrac{x}{\pi^2})(1-\dfrac{x}{2^2\pi^2})(1-\dfrac{x}{3^2\pi^2})...$$

Use Vieta formula

Comparing the coefficients of $x$  in both sides of the equation

$$-\dfrac{x}{3!}=-\dfrac{x}{\pi^2}-\dfrac{x}{2^2\pi^2}-\dfrac{x}{3^2\pi^2}-...$$

Therefore

$$-\dfrac{1}{3!}=-\dfrac{1}{\pi^2}-\dfrac{1}{2^2\pi^2}-\dfrac{1}{3^2\pi^2}-...$$

$$\dfrac{1}{1}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...=\dfrac{\pi^2}{3!}$$

$$\displaystyle\sum_{a=1}^{\infin}\dfrac{1}{a^2}=\dfrac{\pi^2}{6}$$