Absolute convergence of $\sum_{k=1}^{\infty} 8^kk^{2k}(\cos(1/3k)-1)^k$ How do I study the absolute convergence of this series?
I found that the series is divergent using the root test, but idk if it can be absolutely convergent? Is that possible?
$$\sum_{k=1}^{\infty} 8^kk^{2k}(\cos(1/3k)-1)^k$$
Edit: Here is the root test
$$\sqrt[k] {|8^kk^{2k}(\cos(1/3k)-1)^k|}  = |8k^{2}(\cos(1/3k)-1)|$$
$$\lim_{k \to \infty}{|8k^{2}(\cos(1/3k)-1)|}=4/9<1$$

To determine whether the series converges absolutely or not, you have to study the convergence of:
$$\sum_{k=1}^{+\infty} \left| 8^k k^{2k}\ \left( \cos\left(\frac{1}{3k}\right) -1\right)^k \right|$$
Now, since
$$\left| 8^k k^{2k}\ \left( \cos\left(\frac{1}{3k}\right) -1\right)^k \right|=8^k k^{2k}\  \left|\cos\left(\frac{1}{3k}\right) -1 \right|^k $$
applying the root test, as you've done, you have that
$$\sqrt[k]{8^k k^{2k}\left|\cos\left(\frac{1}{3k}\right) -1 \right|^k} \to\frac{4}{9}<1 $$
Hence the series converges absolutely.
