I want to evaluate the following sum
$$ \sum_{n=0}^{\infty}\arctan {(\sin(n ! ))}\, $$ I can't judge if that sum converge or diverge because $~\lim \sin(n!)~$ dosn't exist, however its seems converge according to WA from it's partial sum.
Now I have started my idea by : we have $$\arctan x= x\int_{0}^{1}\dfrac{1}{1+t^2x^2} dt~.$$ In other side we have $$ \sin(x!)=\sum_{n=0}^{\infty} \dfrac{(-1)^{n} x!^{1+2n}}{(2n+1)!}~.$$ Now we have from both definition of $\sin $ and $\arctan $ using series and integral representation we replace each thing in the titled sum , we w'd get :$$\sum\arctan \sin (x!)= x\int_{0}^{1}\dfrac{1}{1+t^2x^2}\sum_{n=0}^{\infty} \dfrac{(-1)^{n} x!^{1+2n}}{(2n+1)!} dt=\sum_{n=0}^{\infty} \dfrac{(-1)^{n} xx!^{1+2n}}{(2n+1)!} \int_{0}^{1}\dfrac{1}{1+t^2x^2}dt~,$$ Then I can't get the next step its complicated, may am not familiar with integral interchange ?