Handling a sequence in a series: $\sum_{n=1}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}$ for $a_n \rightarrow \tfrac{1}{2}$ The question is whether the series $\sum_{n=1}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}$, where  $\{ a_n \}$ is a sequence of positive numbers that converges to ½, converges absolutely or not. 
My hesitation is that we know nothing about the behaviour of $a_n$ (whether it is strictly increasing/decreasing or oscillatory). Here is what I am thinking, but I'm uncertain if it is sufficiently attentive to details:
$\displaystyle \sum_{n=1}^{\infty} \frac{(a_n)^n\cos(n\pi)}{n} = \sum_{n=1}^{\infty} \frac{(a_n)^n(-1)^n}{n}$
We're interested in absolute convergence, so we consider the absolute value of that series, viz: 
$\displaystyle \sum_{n=1}^{\infty} \frac{(a_n)^n}{n}$
Case I: Assume $a_n \rightarrow ½$ from above, then there exists some $n > N$ such that $a_n < r$, where $r<1$, for all $n > N$. As such, 
$\displaystyle \sum_{n=N}^{\infty} \frac{(a_n)^n}{n} < \sum_{n=N}^{\infty} \frac{(r)^n}{n} <   \sum_{n=N}^{\infty} r^n$,
which is a convergent geometric series because $r<1$. 
&nd then similarly for for $a_n \rightarrow ½$ from below. Now that I think of it, we could probably handle the entire thing in one case by using absolute values, right?
 A: An approach.
Let $0<\epsilon<1$. Since $\left\{a_n\right\}_{n\geq1}$ converges to $\frac 12$, then 
$$
\exists \,N\geq0, \forall n\geq N, \left|a_n-\frac 12\right|\leq \epsilon.
$$
Taking for example $\epsilon:=\frac14$,
$$
|a_n|\leq\left|a_n-\frac 12\right|+\frac 12\leq\frac 34, \quad n\geq N,
$$
and
$$
|a_n|^n\leq\left(\frac{3}{4}\right)^{n}, \quad n\geq N.
$$
Then you may write
$$
\left|\sum_{n=N}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}\right|\leq \sum_{n=N}^{\infty} \frac{1}{n}\left(\frac{3}{4}\right)^{n}
$$
But on the right handside we have the rest of a convergent series, thus on the left handside the rest of our series may be as small as we want as $N$ is great.
The initial series is absolutely convergent.
A: We know that $\{a_n\}$ is a sequence of strictly positive numbers that converges to $\frac{1}{2}$, so we can take the absolute value of the whole thing to get rid of the cosine term, and perform the Comparison Test with the series $${\sum_{n=1}^\infty} {a_n}^n$$ which is a geometric series for all $n > N $ for some $N$ and hence converges. We know that: 
$${\sum_{n=1}^\infty} \frac{{a_n}^n}{n} \leq {\sum_{n=1}^\infty} {a_n}^n$$
and the result should fall out. Obviously not the most rigorous proof, and you would need to elaborate on this argument, but that should be the general gist of it. 
