Question

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

Answer 1

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.