# Find out whether $\sum_{k=1}^{\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^{2k}$ converges absolutely, conditionally or diverges.

I need to find out whether the following series converges or diverges. If it converges I also have to discover whether it is an absolute convergence or a conditional one. $$\sum_{k=1}^{\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^{2k}$$ Root test did not help me much since $$\lim_{k\rightarrow\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^2$$ may equal anything
from $$0$$ to $$4$$. Comparison tests also did not work well. I guess this series is convergent, though. But I still have no idea how to prove it.

• $0\leq \frac{1+\cos k}{2+\cos k}\leq\frac{2}{3}$ – Pink Panther Oct 5 '19 at 8:14

The even exponent $$2k$$ guarantees the terms are positive, so convergence cannot be conditional. Since $$1+\cos k\in(0,\,2)\implies\frac{1+\cos k}{2+\cos k}\in(0,\,\frac23)$$, the sequence is bounded above by the geometric progression $$\sum_{k\ge1}(4/9)^k$$, so converges absolutely.
$$\left(\frac{1+\cos k}{2+\cos k}\right)^{2}=\left(1-\frac{1}{2+\cos k}\right)^{2}$$
which is bounded between $$0$$ and $$4/9$$, therefore
$$\left(\frac{1+\cos k}{2+\cos k}\right)^{2k} \le \left(\frac{4}{9}\right)^k$$