$$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^a(\ln n)^3}$$ I want to study the absolute convergence and conditional convergence of this series depending on $a$.
My work:
In class we have been using this convergence test:
$l= \lim_{x\to\infty} \frac{\ln(\frac{1}{a_n})}{\ln n}$ then if $l>1, a_n$ converges, if $l<1, a_n$ diverges and if $l=1$ it is unknown.
Using this and applying absolute value I have:
$$l= \lim_{n\to\infty}\frac{\ln(n^a(\ln n)^3)}{\ln(n)}=\frac{\ln(n^a)+\ln(\ln n )^3)}{\ln n}=\frac{a\ln n+3\ln(\ln n)}{\ln n}= \lim_{n\to\infty}a+\frac{3\ln(\ln n)}{\ln n} $$
$$\lim_{n\to\infty}a+\lim_{n\to\infty}\frac{3\ln(\ln n)}{\ln n}=a+0=a $$ This means that if $a<1$ the series diverges, if $a>1$ the series converges and if $a=1$ we dont know (and should study this case later)
If a series is absolutely converges it also converges, therefore the serie converges and absolutely converges for $a>1$. To finish the problem now I should study for which values of $0<a\leq{1}$ the series converges.
Is this correct? Are there any other easier convergence test I could use here?