# 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 $\sum {\left(- 1\right)}^{n} {a}_{n}$ converges if $| {a}_{n} |$ decreases monotonically and ${\lim}_{n \to \infty} {a}_{n} = 0$.

In this case, we have an alternating series with ${a}_{n} = \frac{1}{\sqrt{\ln \left(n\right)}}$. As $0 < \frac{1}{\sqrt{\ln \left(n\right)}} < \frac{1}{\sqrt{\ln \left(n + 1\right)}}$, we have that $| {a}_{n} |$ is monotonically decreasing, and ${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} \frac{1}{\sqrt{\ln \left(n\right)}} = \frac{1}{\infty} = 0$. Thus, by the alternating series test, the series converges.