Note that the function #sin(n)# oscillates in between #-1lt=sin(n)<=1#. This means that when #sin(n)# is multiplied by a function, its absolute value will be less than or equal to the original function.
In this situation, we see that #abs(sin(n)/(n!))<=1/(n!)#.
By the direct comparison test, if we can show that #1/(n!)# is a convergent series, then #sin(n)/(n!)# will be a convergent series as well since it's less than (or equal to) #1/(n!)#.
We can test the behavior of #sum_(n=1)^oo1/(n!)# using the ratio test.
Applying the ratio test gives:
#lim_(nrarroo)abs((1/((n+1)!))/(1/(n!)))=lim_(nrarroo)abs((n!)/((n+1)!))=lim_(nrarroo)abs(1/(n+1))=0#
Since #lim_(nrarroo)abs(a_(n+1)/(a_n))=0<1#, we know that #sum_(n=1)^oo1/(n!)# is convergent.
Then, as shown before, #sum_(n=1)^oosin(n)/(n!)# is also convergent through the direct comparison test.
