Series n1 to infinity of sine3nnSolutionThe absolute series

Solution

The absolute series |sin(3n)/n| is divergent
Proof:
Let u such that
u>0 and 2u+3<

Then |sin(3n)|=|sin(3n mod 2)|>sin(u) for an infinity of times

If not, suppose there is an m such that |sin(3n)|<sin(u)

then we can assume -u<3n mod 2<u (the other case -u<3n mod 2<+u is similar)

Then u<-u+3<3n+3=3(n+1) mod 2<u+3<-u

so u<3(n+1) mod 2<-u, which implies |sin(3(n+1))|>sin(u)

Since |sin(3n)|>sin(u) an infinity of times, then the series is divergent

The series sin(3n)/n is however convergent

This can be seen by Dirichlet test

since |sin(3)+sin(6)+...+sin(3N)|=|sin(3(N+1)/2)*sin(3N/2)/sin(3/2)|<=1/|sin(3/2)| for any N

So in conclusion the series is conditionally divergent

Sources

http://en.wikipedia.org/wiki/Dirichlet%27s_test

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Other_sums_of_trigonometric_functions