How do you test for convergence of #Sigma (-1)^n/sqrt(lnn)# from #n=[3,oo)#?

1 Answer
Feb 3, 2017

The series converges by the alternating series test.

Explanation:

The alternating series test states that a series #sum(-1)^na_n# converges if #|a_n|# decreases monotonically and #lim_(n->oo)a_n = 0#.

In this case, we have an alternating series with #a_n = 1/sqrt(ln(n))#. As #0 < 1/sqrt(ln(n)) < 1/sqrt(ln(n+1))#, we have that #|a_n|# is monotonically decreasing, and #lim_(n->oo)a_n = lim_(n->oo)1/sqrt(ln(n)) = 1/oo = 0#. Thus, by the alternating series test, the series converges.