Show series [sin(n)]/n converges?

[wisvuze]

nice discussion here; I know the thread is a bit old but since so much trouble was stirred over this, I would like to also point out that dirichlet's test would have worked with this ( and would have proven this fairly easily ). It uses a lemma by Abel which says that if {bn} is non-decreasing and non-negative for each n, and terms a1,..,an are bounded: m <= a1 +... + an <= M for any n , then bk*m <= a1b1 +... + anbn <= bk*M.

Development of these theorems can be found in Spivak's Calculus book on chapters 22 (chapter on infinite series ) and 19 (chapter on integration in elementary terms ). In chapter 22, the Dirichlet test is developed in exercise 13 and in chapter 19, abel's theorem is developed in problem 35.
Not saying that it is very difficult at all, but for anyone who may be curious -- those are excellent sources

 

[myriadbeings]

one other technique that would work is an interesting criterion involving the partial sums. let sn be the nth partial sum of the series. If for every epsilon greater than zero there exists an N such that for all n>N we have
|sn+k - sn|<epsilon for all k >= 1, then the series must converge. (Knopp "Theory and Application of Infinite Series") A quick induction on k would make quick work of this series' convergence.