Does $\sum\limits_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}$ converge? The original question, given to my Calculus II recitation class, was:
Determine if the series $$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}\cos^{2}(n+1)}{n}$$
converges absolutely, conditionally, or diverges. I can kind of see a comparison with the alternating harmonic series here, but making that formal is tough. With the absolute series $\sum\limits_{n=1}^{\infty} \frac{\cos^{2}(n+1)}{n}$, I'm not sure what test to apply.
What I've Tried: No tests (in the classical Calc II curriculum) work. I've tried expanding $\cos^{2}(n+1)$ into a power series within the series in question, but I'm not really sure where to go from there. My intuition tells me this series will diverge, since it seems "close" to the harmonic series; but $\cos(x)$ is less than $1$ infinitely often. 
 A: Regarding the question on absolute convergence, the answer is no.
$$\sum_{n=1}^m\frac{\cos^2 (n+1)}{n} = \underbrace{\sum_{n=1}^m\frac{1}{2n}}_{\text{divergent harmonic series}} + \underbrace{\sum_{n=1}^m\frac{\cos 2(n+1)}{2n}}_{\text{convergent by Dirichlet test}}$$
Also we can use the same approach to prove convergence when $(-1)^n$ appears.
$$\sum_{n=1}^m\frac{(-1)^n\cos^2 (n+1)}{n} = \underbrace{\sum_{n=1}^m\frac{(-1)^n}{2n}}_{\text{convergent alternating series}} - \underbrace{\sum_{n=1}^m\frac{(-1)^{n+1}\cos 2(n+1)}{2n}}_{\text{convergent by Dirichlet test}}$$
Note that $(-1)^{n+1}\cos 2(n+1) = \cos [(n+1)\pi] \cos 2(n+1) = \cos [(n+1)(2 + \pi)]$
A: $\sum_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}$ clearly diverges.  For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $\pi$ by at least $\frac{1}{2}$, which is more than $\frac{\pi}{8}$.  That number of the pair will have a cosine greater than $\cos(\frac{3\pi}{8})$ in absolute value, and $\cos(\frac{3\pi}{8})=\sin(\frac{\pi}{8})>\frac{1}{2}\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{4}>\frac{1}{3}$.  Thus, any two consecutive terms $\frac{\cos^{2}(j+1)}{j} + \frac{\cos^{2}((j+1)+1)}{j+1}$ will contribute at least $\frac{1}{9(j+1)}$ to the sum.  Since $j\ge 1$, we have $\frac{1}{j+1}\ge\frac{1}{4j}+\frac{1}{4(j+1)}$, so $\frac{1}{9(j+1)} \ge\frac{1}{36j}+\frac{1}{36(j+1)}$, i.e. $$\frac{\cos^{2}(j+1)}{j} + \frac{\cos^{2}((j+1)+1)}{j+1} \ge \frac{1}{36j}+\frac{1}{36(j+1)}$$  But this means $$\sum_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}\ge\frac{1}{36}\sum_{n=1}^{\infty}\frac{1}{n}$$ which diverges.
A: You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and
\begin{align}
\left|\sum^N_{n=1}(-1)^n \cos^2(n+1) \right|\leq 1.
\end{align}
Hence the series converges. 
