Show that the series is absolutely convergent The series is
$$\sum^\infty_{n=2} \frac{(-1)^n}{n(\ln(n))^3}$$
I tried the ratio test which did not do anything.  I also tried the root test which gave me
$$\frac{-1}{\sqrt[n]{n}\cdot (\ln(n)^3-n)}$$
which I don't think is right.  Is their another test I can do to confirm that this series is absolutely convergent?  Thanks in advance.
 A: Hint:
Is $$ \sum_{n=2}^{\infty} (-1)^n \frac{1}{n \cdot \ln^3n} $$ convergent or divergent? (Alternating series test)
Is $$ \sum_{n=2}^{\infty} \left |(-1)^n \frac{1}{n \cdot \ln^3n} \right |= \sum_{n=2}^{\infty} \frac{1}{n \cdot \ln^3n} $$  convergent or divergent (Integral test, let $u=\ln x \implies du=\frac{1}{x} dx$ ?
If the above summations are both convergent then $$ \sum_{n=2}^{\infty} (-1)^n \frac{1}{n \cdot \ln^3n} $$ is absolutely convergent. 
A: $\textbf{Integral Test}$
$$\sum_{n=2}^{\infty} \frac{1}{n (\ln(n))^3} \ \ \ \text{converges} \iff \lim_{R\to \infty} \int_{2}^{R} \frac{1}{x(\ln(x))^3}\ \text{dx}\ \ \text{converges}$$
Let $u =\ln(x), du = \frac{1}{x} \ \text{dx}\ $ ... (go from here!)
A: Hint: Absolute convergence of
$$
\sum_{n=1}^\infty a_n
$$
is equivalent to convergence of
$$
\sum_{n=1}^\infty|a_n|
$$
Thus, absolute convergence of the series given is equivalent to convergence of
$$
\sum_{n=1}^\infty\frac1{n\log(n)^3}
$$
A good test to try when logs are involved is the Cauchy Condensation Test. This would mean testing whether
$$
\sum_{n=1}^\infty2^n\frac1{2^n(n\log(2))^3}
=\frac1{\log(2)^3}\sum_{n=1}^\infty\frac1{n^3}
$$
converges.
