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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.

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

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We have that

$$\left(\frac{1+\cos k}{2+\cos k}\right)^{2}=\left(1-\frac{1}{2+\cos k}\right)^{2}$$

which is bounded between $0$ and $4/9$, therefore

$$\left(\frac{1+\cos k}{2+\cos k}\right)^{2k} \le \left(\frac{4}{9}\right)^k$$

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