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!
 A: For the first - don't only state that $|\frac{1}{\ln(n)}|$ is decreasing - the fact that $\lim_{n\to\infty}|\frac{1}{\ln(n)}|=0$ is certainly relevant.
The alternating series test states:

if $|a_n|$ decreases monotonically, all $a_n$ are positive or all negative, and $\lim_{n\to\infty}a_n$, then the alternating series $a_0-a_1+a_2-\cdots$ converges.

It is not enough to say $|a_n|$ is decreasing, nor is it to say that $|a_n|$ is monotonically decreasing and bounded. Take for example $a_n=1+\frac{1}{n}$, which is monotonically decreasing, and certainly bounded. But the alternating series $a_0-a_1+a_2-a_3+\cdots$ definitely doesn't converge - since each $a_n$ is about 1 for large $n$, this series keeps bouncing back and forth (this is obviously not a formal proof that it doesn't converge; I hope I made my point clear though).

For the second, probably the sequence is meant rather than the series; $(a_n)_{2}^{\infty}$ is common notation for the sequence $(a_2,a_3,a_4,...)$, and as such, the question is, does the sequence $(\frac{1}{\ln(2)},\frac{1}{\ln(3)},\frac{1}{\ln(4)},...)$ converge? As you noted, $\ln(n)$ goes to infinity as $n$ approaches infinity - thus, $\frac{1}{\ln(n)}$ approaches $0$ as $n$ approaches infinity, and therefore the sequence converges.

You only need the comparison test for the third one - simply note that $|\cos(\pi n/100)|\leq 1$. Can you now bound $\sum_{n=1}^{\infty}2^{-n/4}\cos(\pi n/100)$ by something you do know converges?

As you might know, $|a+b|\leq |a|+|b|$. We can use this fact on the above series to note
$$\left|\sum_{n=1}^{\infty}2^{-n/4}\cos(\pi n/100)\right|\leq\sum_{n=1}^{\infty}\left|2^{-n/4}\cos(\pi n/100)\right|$$
and we know that $|2^{-n/4}\cos(\pi n/100)|=|2^{-n/4}||\cos(\pi n/100)|\leq 2^{-n/4}$ and thus,
$$\left|\sum_{n=1}^{\infty}2^{-n/4}\cos(\pi n/100)\right|\leq\sum_{n=1}^{\infty}2^{-n/4}$$
Now the only thing left for you to do is find a bound or value to $\sum_{n=1}^{\infty}2^{-n/4}$.
A: 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.
