Does #a_n=sin(n)/n # converge? Calculus Tests of Convergence / Divergence Infinite Sequences 1 Answer Konstantinos Michailidis May 28, 2016 Because #-1<=sinn<=1# hence #-1/n<=sinn/n<=1/n# hence #1/n->0# as n goes to infinity then #0<=sinn/n<=0# as n goes to infinity hence the sequence converges. Answer link Related questions What is the difference between an infinite sequence and an infinite series? What is the definition of an infinite sequence? How do you Find the limit of an infinite sequence? How do you Find the #n#-th term of the infinite sequence #1,1/4,1/9,1/16,…#? How do you Determine whether an infinite sequence converges or diverges? How do you determine whether the infinite sequence #a_n=(2n)/(n+1)# converges or diverges? How do you determine whether the infinite sequence #a_n=(1+1/n)^n# converges or diverges? How do you determine whether the infinite sequence #a_n=(-1)^n# converges or diverges? How do you Find the #n#-th term of the infinite sequence #1,-2/3,4/9,-8/27,…#? How do you determine whether the infinite sequence #a_n=e^(1/n)# converges or diverges? See all questions in Infinite Sequences Impact of this question 35291 views around the world You can reuse this answer Creative Commons License