I'm having a lot of trouble with this section on series, as many of you who viewed and responded (thanks!) to my other posts know. I can't do about 20 of the problems provided, out of 30, so I think I'm missing something "integral" here.

I re-read the chapter, and I think that perhaps what I'm missing is the proper use of Dirichlet's and Abel's test. It seems, from my reading, that these two tests are really pretty much equivalent. I present Abel's test, as it is written in my book, below:

Let $\displaystyle \sum a_n$ be a covergent series of complex terms, and let $\displaystyle \{b_n\}$ be a monotonic convergent sequence of real terms. Then the series $\displaystyle \sum a_n b_n$ converges.

Here's a few example problems that I'm stuck on. I'm not presenting a few because I want/expect answers on how to solve all of them, but rather I am presenting these to see if Abel/Dirichlet's tests would even be appropriate for them:

$\displaystyle \sum_{n=1}^\infty \log \left(n \sin \frac 1 n \right)$

$\displaystyle \sum_{n=1}^\infty (-1)^n \left ( 1 - n \sin \frac 1 n \right )$

$\displaystyle \sum_{n=1}^\infty (-1)^n \left ( 1 - \cos \frac 1 n \right )$

$\displaystyle \sum_{n=1}^\infty (-1)^n \arctan \frac 1 {2n+1}$

$\displaystyle \sum_{n=1}^\infty (-1)^n \left ( \frac \pi 2 - \arctan(\log n) \right )$

OK, that's probably enough to get an idea of the problems that I'm struggling with. The question for each of these is if they are absolutely convergent. From the way that Abel's test is presented, I thought that it would be most useful if you had a series which was presented in as a product, similar to integration by parts, and you knew that one of those parts was convergent. None of these seem to have that form, however I attempted to force it into this form with the following example:

$\displaystyle 1 - n \sin \frac 1 n=\frac{1}{n^2} \left(n^2-n^3 \sin \frac 1 n \right)$

$\displaystyle \frac {d} {dn} \left(n^2-n^3 \sin \frac 1 n \right) = 2n-3n^2 \sin \frac 1 n +n \sin \frac 1 n=n(2-3n\sin\frac 1 n)+n\sin\frac 1 n$

Now since $\displaystyle \lim_{n \to \infty} n \sin \frac 1 n = 1$, the derivative is always negative for all $\displaystyle n >N$ for some N. Therefore, we can use the convergence of $\displaystyle \sum \frac 1 {n^2}$ and Abel's test to conclude that the series $\displaystyle \sum (-1)^n \left ( 1 - n \sin \frac 1 n \right)$ is absolutely convergent.

Is this the correct approach to this type of problem? How would this be applicable to, say, $\displaystyle \sum \lim \left (n\sin\frac 1 n \right)$? Thanks in advance.