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

Question

Emilia Aguilar · Accepted Answer

The series converges by the alternating series test.

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.

Isaiah Allen · Answer

undefined To test for the convergence of the series $ \sum_{n=3}^{\infty} \frac{(-1)^n}{\sqrt{\ln n}} $, you can use the alternating series test. This test states that if a series is alternating, meaning that its terms alternate in sign, and the absolute value of the terms decreases as $ n $ increases, and the terms approach zero as $ n $ approaches infinity, then the series converges.

In this series, the terms alternate in sign and approach zero as $ n $ approaches infinity. To check if the terms decrease in absolute value, you can observe the behavior of the function $ \frac{1}{\sqrt{\ln n}} $ as $ n $ increases.

Since $ \ln n $ increases as $ n $ increases, $ \sqrt{\ln n} $ also increases. Therefore, $ \frac{1}{\sqrt{\ln n}} $ decreases as $ n $ increases.

Because the series meets all the conditions of the alternating series test, it converges.