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.

  • $\begingroup$ Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious $\endgroup$ – imranfat May 6 '18 at 5:49
  • $\begingroup$ @imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me. $\endgroup$ – Fred May 6 '18 at 5:52
  • $\begingroup$ Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious... $\endgroup$ – imranfat May 6 '18 at 16:11

$\sum_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}$ 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.


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 $(-1)^{n+1}\cos 2(n+1) = \cos [(n+1)\pi] \cos 2(n+1) = \cos [(n+1)(2 + \pi)]$

  • $\begingroup$ Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion. $\endgroup$ – Fred May 6 '18 at 5:41
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    $\begingroup$ @Fred. Yes you can split the partial sums as I added above. As $m \to \infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges. $\endgroup$ – RRL May 6 '18 at 5:45
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    $\begingroup$ This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $\sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below. $\endgroup$ – C Monsour May 6 '18 at 5:49
  • $\begingroup$ @Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) \to 0$ in a decreasing fashion and the partial sums $\sum_{n=1}^{m} \cos 2(n+1)$ are bounded for all $m$. Same applies to $\sum_{n=1}^m \cos ((2+\pi)(n+1)$. $\endgroup$ – RRL May 6 '18 at 6:06

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.

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    $\begingroup$ True, but this test is not in Calculus II curriculum. $\endgroup$ – imranfat May 6 '18 at 5:45
  • $\begingroup$ @CMonsour Yes. But I didn't say $\sum^N_{n=1}\cos^2(n+1)$ is bounded. Look carefully. $\endgroup$ – Jacky Chong May 6 '18 at 5:48
  • $\begingroup$ OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove. $\endgroup$ – C Monsour May 6 '18 at 5:55
  • $\begingroup$ Jacky.Very nice, clear and concise.Thanks! $\endgroup$ – Peter Szilas May 6 '18 at 6:51

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