How to figure out either series are absolute or conditional convergence $\sum_1^\infty \cos(\frac{\pi n}{3})(n^{\frac{1}{\sqrt[6]{n+6}}}-1)$ $$\sum_1^\infty \cos\left( \frac{\pi n} 3 \right) \left(n^{\frac 1 {\sqrt[6]{n+6}}}-1\right)$$
I tried to use Dirichlet, but unsuccessfully. Please, give me hints. Thanks a lot.
 A: Convergence is not absolute. Since $x>0\implies e^x>1+x $, we have $$n>1\implies  n^{(n+6)^{-1/6}}-1=e^{(n+6)^{-1/6}\ln n}-1>(n+6)^{-1/6}\ln n.$$
Let $y_n$ be the $n$th term of the series. For $k\in \mathbb N$  we have $\cos 
 \pi (6k+1)/3=1/2,$  so we have $$y_{(6k+1)}>(1/2)(6k+7)^{-1/6}\ln (6k+1)>(1/2)(6^6k^6)^{-1/6}\ln (6k)=(1/2)(\ln 6k)/(6k). $$
$$\text {Therefore }\quad  \sum_{n=1}^{\infty}|y_n|\geq \sum_{k=1}^{\infty}|y_{(6k+1)}|\geq (1/2)\sum_{k=1}^{\infty} (\ln 6k)/6k\geq (1/12)\sum_{k=1}^{\infty}1/k=\infty.$$
A: \begin{align}
& \Big( \cos \frac{1\pi} 3, \quad \cos\frac{2\pi} 3, \quad \cos\frac{3\pi} 3, \quad \cos\frac{4\pi} 3, \quad \cos \frac{5\pi} 3, \quad \cos\frac{6\pi} 3, \quad \ldots \Big) \\[10pt]
= \frac 1 2 & \Big( \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \ldots\ldots\ldots  \Big)
\end{align}
The sequence $1,1,0,-1,-1,0$ repeats forever.
So you have a sum of two positive terms, then a sum of two negative terms, etc.
The alternating series test says that if the terms decrease and approach $0$ and have alternating signs, then the series converges.
Whether it converges absolutely takes more work to answer after that. Maybe I'll be back.
