The Series of Pi

By Maclaurin Series ,

$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \ldots$

Let $x = \pi$

$\begin{aligned} \sin(\pi) = 0 & = & \pi - \frac {\pi^3}{3!} + \frac {\pi^5}{5!} - \frac {\pi^7}{7!} + \ldots \\ \pi & = & \frac {\pi^3}{3!} - \frac {\pi^5}{5!} + \frac {\pi^7}{7!} - \ldots \\ 1 & = & \frac {\pi^2}{3!} - \frac {\pi^4}{5!} + \frac {\pi^6}{7!} - \ldots \\ \end{aligned}$

Note that $\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots$ , thus our desired sum is $1 - \frac{\sin \pi}{\pi} = 1$ .

Subtract one on the left , leave the +1 unchanged , and insert the negative one inside the series then take 1/pi common factor

we have : 1 - (1/pi) [ pi - pi^3/3! + pi^5/5! - .... ] which is 1 - (1/pi) sin pi , sin pi = o , we get the answer = 1

Don't focus on the $\pi^{2n}$ , focus on the $(2n+1)!$

be suspicious that mac laurin series of $sin x$ somehow must come up

by a little inspection, the series is equal to $1- \frac{sin x}{x}$ evaluated at $x = \pi$