Day: April 17, 2023

Deriving the Extraordinary Euler Sum

/ Michael Xue / Leave a comment

By definition,

\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} + \dots

Dividing by x \ne 0 yields

\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}+\dots\quad\quad\quad(1)

Assuming the properties of finite polynomials hold true for infinite series,

\sin(x) = 0 when x=\pm \pi, \pm 2\pi, \pm 3\pi, ...

means

1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}+\dots

= (1-\frac{x}{\pi})(1-\frac{x}{-\pi})(1-\frac{x}{2\pi})(1-\frac{x}{-2\pi})(1-\frac{x}{3\pi})(1-\frac{x}{-3\pi})\dots

= (1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})(1-\frac{x}{3\pi})(1+\frac{x}{3\pi})\dots

=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})(1-\frac{x^2}{16\pi^2})\dots

i.e.,

1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}+\dots=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})(1-\frac{x^2}{16})\dots\quad\quad\quad(2)

It can be shown (see Exercise-1) that the coefficient of x^2 on the right hand side of (2) is

-\frac{1}{\pi^2}\left(1+ \frac{1}{4}+ \frac{1}{9} + \frac{1}{16}+\dots\right).

Therefore, equate the like powers of x^2 on both sides of (2) gives

-\frac{1}{3!} = -\frac{1}{\pi^2}\left(1+ \frac{1}{4}+ \frac{1}{9} + \frac{1}{16} + \dots\right).

Multiplying -\pi^2 throughout, we obtain the Extraordinary Euler Sum

\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots

Omega knows Euler’s sum

Exercise-1 show the coefficient of x^2 on the right hand side of (2) is

-\frac{1}{\pi^2}\left(1+ \frac{1}{4}+ \frac{1}{9} + \frac{1}{16} + \dots\right).

Exercise-2 Show that \sum\limits_{i=1}^{n} \frac{1}{i^2} is convergent.