How do you test the alternating series #Sigma (-1)^nsqrtn# from n is #[1,oo)# for convergence? Calculus Tests of Convergence / Divergence Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series 1 Answer Andrea S. Jun 8, 2018 As #lim_(n->oo) sqrtn = +oo# the series does not satisfy Cauchy's necessary condition and thus cannot be convergent. Besides let #a_n = (-1)^n sqrtn#. Clearly #a_(2k) > 0# and #a_(2k+1) < 0# so the series oscillates indefinitely. Answer link Related questions What is the Alternating Series Test of convergence? Can the Alternating Series Test prove divergence? Does the Alternating Series Test determine absolute convergence? How do you use the Alternating Series Test? What do you do if the Alternating Series Test fails? What is an example of an alternating series? How do I find the sum of the series: 4+5+6+8+9+10+12+13+14+⋯+168+169+170. since D is changing... How do you determine the convergence or divergence of #sum_(n=1)^(oo) (-1)^(n+1)/n#? How do you determine the convergence or divergence of #Sigma ((-1)^(n+1)n)/(2n-1)# from #[1,oo)#? How do you determine the convergence or divergence of #Sigma ((-1)^(n+1))/(2n-1)# from #[1,oo)#? See all questions in Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series Impact of this question 1902 views around the world You can reuse this answer Creative Commons License