How to evaluate $\sum_{n=0}^\infty {\frac {\sin \left( n! \right) }{n!}}$

Question

Have you got any ideas on how to evaluate this serie ?

$$\sum_{n=0}^\infty {\frac {\sin \left( n! \right) }{n!}}$$

It's convergent and is about ${2.057545430}$, but I don't find any ways to get a closed form.

Answer 1

I am skeptical about a possible closed form but you can always represent the result as (for example) $$\sum_{n=0}^\infty {\frac {\sin \left( n! \right) }{n!}}\sim a \pi$$ where tha $a$ is a rational number.

For example $a=\frac{3470614198}{5299156903}$ would give a absolute relative error of $2.79 \times 10^{-18}$.

Interesting enougth is also the fact that the number can be approximated as the product of $\pi$ by the first positive root of $$6074 x^4-6882 x^3+2117 x^2-7479 x+4806=0$$ which would give a absolute relative error of $1.36 \times 10^{-19}$.

Edit

For the solution of $$\sum_{n = 0}^\infty \frac{\sin(n)}{n!}$$ given by @WhatsUp in comments, consider $$\Im\left(\sum_{n = 0}^\infty \frac{\left(e^i\right)^n}{n!}\right)=\Im\left(e^{e^i}\right)=\Im\left(e^{\cos(1)+i\sin(1)}\right)=e^{\cos (1)}\Im\left( \cos (\sin (1))+i \sin (\sin (1)) \right)$$