Find convergence of series .... How do we find the convergence of $$ s_n = \sin (1!) +  \frac {\sin (2!)}{1!} + \frac {\sin (3!)}{2!} + \frac {\sin (4!)}{3!} +... \frac {\sin (n+1!)}{n!} $$ 
I was thinking of using convergence tests but they are used only for infinite series.  I tried to use the ratio test $ \frac {a_{(n+1)}}{a_n}$ but the series does not go to infinite so I couldn't use it. Meanwhile I also tried to simplify it, but still I couldn't get something concrete.

$$\frac{\sin(n+1)!}{n!}\leq \frac{1}{n!}\\S_{n}=\sin(1!) +\frac{\sin(2!)}{1!} +\frac{\sin(3!)}{2!} +
\frac{\sin(4!)}{3!} +...+\frac{\sin(n+1)}{n!} \\\leq |\sin(1!) +\frac{\sin(2!)}{1!} +\frac{\sin(3!)}{2!} +\frac{\sin(4!)}{3!} +...+\frac{\sin(n+1!}{n!}| \\\leq |\sin(1!)| +|\frac{\sin(2!)}{1!}| +|\frac{\sin(3!)}{2!}| +|\frac{\sin(4!)}{3!}\ +...+|\frac{\sin(n+1)!}{n!}|\\\leq \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}\\\leq 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+... =e^1$$ So, it converges.
