# How do you determine the convergence or divergence of Sigma ((-1)^(n))/(ln(n+1)) from [1,oo)?

Feb 2, 2017

The series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} / \ln \left(n + 1\right)$

is convergent.

#### Explanation:

The series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} / \ln \left(n + 1\right)$

is an alternating series, so we can test its convergence using Leibniz's theorem, which states that an alternating series

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$

is convergent if:

(i) ${\lim}_{n \to \infty} {a}_{n} = 0$

(ii) ${a}_{n + 1} \le {a}_{n}$

in our case:

${\lim}_{n \to \infty} \frac{1}{\ln} \left(n + 1\right) = 0$

and since $\ln x$ is a monotone growing function:

$\ln n < \ln \left(n + 1\right) \iff \frac{1}{\ln} n > \frac{1}{\ln} \left(n + 1\right)$

also the second condition is satisfied.