Infinitly summed $\pi$

Using Maclaurin series as follows.

$\begin{aligned} x \sin x & = \frac {x^2}{1!} - \frac {x^4}{3!} + \frac {x^6}{5!} - \cdots \\ \color{#3D99F6} \int x \sin x \ dx & = \frac {x^3}{1!3} - \frac {x^5}{3!5} + \frac {x^7}{5!7} - \cdots & \small \color{#3D99F6} \text{By integration by parts} \\ - x \cos x + \int \cos x \ dx & = \frac {x^3}{1!3} - \frac {x^5}{3!5} + \frac {x^7}{5!7} - \cdots \\ - x \cos x + \sin x + \color{#3D99F6} C & = \frac {x^3}{1!3} - \frac {x^5}{3!5} + \frac {x^7}{5!7} - \cdots & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \implies C & = 0 & \small \color{#3D99F6} \text{Putting }x = 0 \\ \implies - x \cos x + \sin x & = \frac {x^3}{1!3} - \frac {x^5}{3!5} + \frac {x^7}{5!7} - \cdots & \small \color{#3D99F6} \text{Putting }x = \pi \\ \pi & = \frac {\pi^3}{1!3} - \frac {\pi^5}{3!5} + \frac {\pi^7}{5!7} - \cdots \end{aligned}$

$\displaystyle \implies \frac 1{2\pi}\left(\frac {\pi^3}{1!3} - \frac {\pi^5}{3!5} + \frac {\pi^7}{5!7} - \cdots \right) = \frac 1{2\pi} \times \pi = \frac 12$

$\implies mn = 1\times 2 = \boxed{2}$

$\begin{aligned}\text{S} & = \frac{1}{2\pi}\left(\frac{\pi^3}{1!3}-\frac{\pi^5}{3!5}+\frac{\pi}{5!7}+\cdots\right) \\& =\frac{1}{2\pi}\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\pi^{2n+1}}{(2n-1)!(2n+1)} \\& = \frac{1}{2\pi}\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\pi^{2n+1}(2n)}{(2n)(2n-1)!(2n+1)}\\&=\frac{1}{2\pi}\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\pi^{2n+1}(2n+1-1)}{(2n+1)!}\\& =\frac{1}{2\pi} \displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n-1}\pi^{2n+1}(2n+1)}{(2n+1)!}-\frac{(-1)^{n-1}\pi^{2n+1}}{(2n+1)!}\right)\\&= \frac{1}{2\pi} \displaystyle\sum_{n=1}^{\infty}\left((\pi)\frac{(-1)^{n-1}\pi^{2n}(2n+1)}{(2n)!}-\frac{(-1)^{n-1}\pi^{2n+1}}{(2n+1)!}\right) \\& = \frac{\pi}{2\pi}\left({\color{#3D99F6}\frac{\pi^2}{2!}-\frac{\pi^4}{4!}+\frac{\pi^6}{6!}+\cdots}\right)-\frac{1}{2\pi}\left({\color{#D61F06}\frac{\pi^3}{3!}-\frac{\pi^5}{5!}+\frac{\pi^7}{7!}+\cdots}\right) \\& = \frac{1}{2}(1-\cos{\pi})-\frac{1}{2\pi}(\pi-\sin{\pi}) \\& = 1-\frac{1}{2}\\& = \frac{1}{2}\implies\frac{m}{n} \end{aligned}$ $\implies mn = \boxed{2}$