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 2038 views around the world You can reuse this answer Creative Commons License