Question

`sum_(n=1)^oo sin(n)/(n!)` How would i find if it converges or diverges?

Answer 1

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.