Solve this!!
∞∑n=0 (sin2nө/n!)
I'm very rusty on this, but assume we have a function $\displaystyle f(x)$ such that the Fourier sine series for f is
$\displaystyle F[f(x)] = \sum_{n = 0}^{\infty}\frac{1}{n!}sin(2n \theta)$
Then we need to invert the Fourier sine series, which if I remember correctly means that
$\displaystyle f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \sum_{n = 0}^{\infty}\frac{1}{n!}sin(2n\theta) sin(x \theta) d \theta$
Someone needs to check me on that result, but it's going to be something of the sort. (The best way for you, kamaksh_ice, to check me is to do the problem then take the Fourier sine series of the result and see if you get your original series back.)
-Dan