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

Question

Emily Ammerman · Accepted Answer

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.

Charles Arocha · Answer

undefined To determine if the series $ \sum_{n=1}^\infty \frac{\sin(n)}{n!} $ converges or diverges, you can use the Ratio Test.

The Ratio Test states that for a series $ \sum_{n=1}^\infty a_n $, if $ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1 $, then the series converges; if $ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| > 1 $, then the series diverges; and if $ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1 $, the test is inconclusive.

Applying the Ratio Test to the given series, we have:

$$ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{\frac{\sin(n+1)}{(n+1)!}}{\frac{\sin(n)}{n!}}\right| = \lim_{n \to \infty} \left|\frac{\sin(n+1)}{\sin(n)} \cdot \frac{n!}{(n+1)!}\right| $$

Simplify this expression and take the limit. If the limit is less than 1, the series converges; if it's greater than 1, the series diverges; if it's equal to 1, the test is inconclusive.