How do you prove the convergence or divergence of this? Assuming you use Alternating Series, how do I prove this is convergent or divergent? 
$$\sum_{n=8}^{\infty} (-1)^n\frac{(\ln(n))^2}{n}. $$
 A: Since $\dfrac{(\log(n))^2}{n}$ is positive, decreasing and goes to $0$, you can use Dirichlet's test (or the alternating series test) to show that the series converges. This is all what you need to check.
A: To apply the Alternating Series Test, you need to verify, in addition to the obvious alternation of sign, that (i) (ultimately) the terms are decreasing and (ii) the limit of the terms is $0$.
(i) To show the decreasing part, one way is to look at the function $f(x)=\frac{(\log x)^2}{x}$. Calculate $f'(x)$, and show that it is negative for $x\ge 8$.
(ii) There are various ways to show that the limit of the terms is $0$. For example, one can use L'Hospital's Rule to find $\lim_{x\to\infty}\frac{(\log x)^2}{x}$.  
A: In fact, we can evaluate this exactly in terms of well known constants.  While this is beyond the scope of your question, it may interest other readers so I will post this (otherwise too long comment) as an answer.
Specifically, we have that 
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\log^2 n}{n}=\frac{\log^3 2}{3}-2\gamma \log 2-\gamma_1 \log^2 2, $$ where $\gamma$ is the Euler-Mascheroni Constant, and $\gamma_1$ is the first Stieltjes constant.   
See this Math Stack Exchange answer, or this blog post, for more details.
