Are the following convergent or divergent? Are the following convergent or divergent? Justify
$1.\sum_{n=2}^\infty (-1)^n\frac{1}{\ln(n)}$
$2. \left(\frac{1}{ln(n)}\right)^\infty _{n=2}$
$3.\sum_{n=1}^\infty 2^{-n/4}\cos(\pi n/100)$
My thoughts are:
$1.$ is covergent because for $\sum_{n=2}^\infty \frac{1}{\ln(n)}$, $  \lvert\frac{1}{\ln(n)}\rvert$ is decreasing as $n\to\infty$, so by the Alternating series test, $\sum_{n=2}^\infty (-1)^n\frac{1}{\ln(n)}$ converges.
$2.$ is divergent because $n>\ln(n) \implies \frac{1}{n}<\frac{1}{\ln(n)}. $
Because $ \left(\frac{1}{n}\right)^\infty _{n=2}$ diverges and $\frac{1}{n}<\frac{1}{\ln(n)}$. So, by the comparison test, $ \left(\frac{1}{\ln(n)}\right)^\infty _{n=2}$ diverges.
Is what I've written so far true?
My most trouble comes from 
$3.$ 
I think it converges to $0$, but I'm not sure under what tests it does.
Thanks a lot!

The first looks good.
The second looks good if you meant to take $\sum^\infty _{n=2}\frac{1}{\ln(n)}$, but as written, its just a sequence:
$\left(\frac{1}{\ln(n)}\right)^\infty _{n=2}\to0$
The third may be handled with absolute convergence, noting that
$$|\cos(x)|\le1$$
Thus, we are left with
$$\sum_{n=1}^\infty2^{-n/4}$$
Which is a geometric series and converges.
