How can I find if $\sum_{n=2}^\infty {(-1)^n*4 \over (ln(n))^2} $ converges or diverges using the alternating series test? 
$$4\sum_{n=2}^\infty {(-1)^n \over (\ln(n))^2} $$

If $4 \over (\ln(n))^2$ = $u_n$, then $u_n$> 0, and: 
$$\lim_{n\to\infty} u_n = 0 $$
But there is one more test to prove convergence which says that the positive $u_n$'s are nonincreasing, $u_n > u_{n+1}$, and I'm not really sure how to prove that..if someone could lead me in the right direction it'd be much appreciated.
 A: You don't even need calculus to prove that $u_n = \frac{4}{(\ln(n))^2}$ is monotonically decreasing, for large enough $n$.
The argument rests on three facts: 


*

*that $\ln(n)$ is increasing; that is, $\ln(n) < \ln(n + 1)$, that

*for positive $a, b$ both greater than $1$, if $a > b$ then $a^2 > b^2$

*that for positive $a, b, c$, the inequality $\frac{a}{b} > \frac{a}{b + c}$; that "increasing the denominator makes the fraction smaller."
So, once you find the particular $N$ so that $\ln(n) > 1$ for all $n > N$, you can use the inequalities above to show that $u_n > u_{n + 1}$.
Of course, you can use calculus to show that $u_n$ is monotonically decreasing for large enough $n$, but you don't have to.
A: The Alternating Series Test basically says if a series is monotonically decreasing (absolute value of terms is decreasing) and $\lim_{n\to \infty}$ of the series is equal to $0$, then the series converges. 
You are going in the right direction. $$\frac{4}{(\ln n)^{2}}$$ 
decreases monotonically and, as you noted, $\lim_{n\to \infty}$ of this series equals $0$.
Therefore, by the Alternating Series Test, this series converges.
A: One way to prove it is strictly decreasing is to use the first derivative test by considering the function

$$ f(x) = \frac{1}{\ln^2(x)}. $$

A: You have to prove that:
$$\frac{1}{(\ln n)^2}\gt\frac{1}{(\ln (n+1))^2}$$
The inequality is very simple.
