Does #a_n=(n + (n/2))^(1/n) # converge? Calculus Tests of Convergence / Divergence Infinite Sequences 1 Answer Cesareo R. Nov 28, 2016 The sequence is convergent. Explanation: #a_n=(n+n/2)^(1/n)=n^(1/n)(3/2)^(1/n)# so #lim_(n->oo)(n+n/2)^(1/n)=lim_(n->oo)n^(1/n)lim_(n->oo)(3/2)^(1/n)# but #lim_(n->oo)n^(1/n)=1# and #lim_(n->oo)(3/2)^(1/n)=1# so #lim_(n->oo)a_n=1# and 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 2397 views around the world You can reuse this answer Creative Commons License