Does #a_n={(3/n)^(1/n)} #converge? If so what is the limit? Calculus Tests of Convergence / Divergence Infinite Sequences 1 Answer Jim H Mar 27, 2016 It converges to #1#. Explanation: #f(x) = (3/x)^(1/x) = e^(ln(3/x)/x)# #lim_(xrarroo)(ln(3/x)/x)# has initial form #(-oo)/oo# which is indeterminate, so use l'Hospital's rule. #lim_(xrarroo) (x/3(-3/x^2))/1 = lim_(xrarr0)-1/x = 0# Since the exponent goes to #0#, we have #lim_(xrarroo)(ln(3/x)/x) = e^0 = 1# 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 7268 views around the world You can reuse this answer Creative Commons License