# 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:

1. 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).

2. 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.

• The $\frac14$ is completely moot; $\sum_n s_n$ converges (absolutely, uniformly, etc.) iff $\sum_n Cs_n$ converges (absolutely, uniformly, etc.) for any (non-zero!) constant $C$. Nov 18, 2016 at 22:52
• This proof seems to be more work than necessary. Note $$\sum_{k=0}^K \frac{\cos^2(k)}{k+1} =\frac12 \sum_{k=0}^{K} \frac{1}{k+1}+\frac12 \sum_{k=0}^{K}\frac{\cos(2k)}{k+1}$$Note that the second series on the right-hand side converges. Therefore, if the sum on the left-hand side converges, then the harmonic series converges. Nov 18, 2016 at 23:56

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

$$\sum_{k=0}^{\infty}\frac{\cos^2k}{k+1} \ge \sum_{k=0}^{\infty} \frac{m}{k+2} = \infty.$$

• Nicely done. +1 Nov 19, 2016 at 0:23
1. 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 $$\sum_{k=0}^{\infty}\frac{\sin^2{2k}}{2k+1}$$
2. 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.

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.

• This is the better approach. +1 Nov 18, 2016 at 23:22