Prove that $\sum_\limits{k=0}^{\infty}\frac{\cos^2k}{k+1}$ diverges Prove that the series $$\sum_{k=0}^{\infty}\frac{\cos^2k}{k+1}$$ diverges. 
I am told to use the following approach: 
Suppose that the above series converges and then conclude that under this hypothesis, $$\sum_{k=0}^{\infty}\frac{\sin^2{2k}}{2k+1}$$ also converges and find a contradiction.
The first part of the mark scheme uses a chain of inequalities which eventually results in the following:
$a_k \geq \frac{1}{4} \frac{\sin^2{2k}}{2k+1} := b_{2k}$ (where $a_k=\frac{cos^2k}{k+1}$ )
Then, it states: "therefore by the comparison test, the series $$\sum_{k=0}^{\infty}b_{2k} = \sum_{k=0}^{\infty}\frac{\sin^2{2k}}{2k+1}$$ converges absolutely." 
I dont get why the above is true, because:


*

*I fail to see the comparison test in action. The above just shows that $a_k$ is bigger than a fourth of the desired sequence. How can this imply convergence? (I may be neglecting some constant in front, but in the definition given in class, there was no constant provided).

*Since when does $$\sum_{k=0}^{\infty}b_{2k} = \sum_{k=0}^{\infty}\frac{\sin^2{2k}}{2k+1}$$ ??? (where did the $1/4$ go?, or is this just a notation issue and $b_{2k}$ was defined without the $1/4$?)
If someone could clarify for me these (probably trivial) issues, I would be grateful.
 A: Different approach: Note that $\cos^2x + \cos^2(x+1)$ is continuous, never $0,$ and is periodic with period $2\pi.$ It follows that $\cos^2x + \cos^2(x+1)$ has a minimum value $m>0$ on $\mathbb R.$ Thus
$$\frac{\cos^2k}{k+1} + \frac{\cos^2(k+1)}{k+2} \ge \frac{\cos^2k+\cos^2(k+1)}{k+2} \ge \frac{m}{k+2}.$$
for all $k.$ Thus summing in pairs shows
$$\sum_{k=0}^{\infty}\frac{\cos^2k}{k+1} \ge \sum_{k=0}^{\infty} \frac{m}{k+2} = \infty.$$
A: *

*Then you get convergence of
\begin{align}
\sum_{k=0}^{\infty}\frac{1}{4}\frac{\sin^2{2k}}{2k+1}=\frac{1}{4}\sum_{k=0}^{\infty}\frac{\sin^2{2k}}{2k+1}
\end{align}
which implies the convergence of
$$
\sum_{k=0}^{\infty}\frac{\sin^2{2k}}{2k+1}
$$

*You are right,
$$
\sum_{k=0}^{\infty}b_{2k} = \sum_{k=0}^{\infty}\frac{1}{4}\frac{sin^2{2k}}{2k+1}
$$
but as noted in 1. this is enough for what you want.

A: Starting from
$$\frac{\cos(2k)}{k+1}+\frac{1}{k+1}=2\frac{\cos^2(k)}{k+1}.$$
observe that
$$\sum \frac{\cos(2k)}{k+1}$$
converges  by Abel's criterion,
and
$\sum \frac{1}{k+1}$ diverges as an harmonic series,
thus 
convergent  $+$ divergent $=$ divergent.
