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Study the convergence of $\sum_{n=1}^\infty \frac{(-1)^n \cos^2(n)}{n}$

Abel's/Dirichlet's tests cannot be applied here.
I guess it's something more tricky involving integration maybe (?)

AlonAlon
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  • What is the maximum outcome of $cos^2n$ and what do you know about the alternating harmonic series? – imranfat Jan 26 '15 at 16:06
  • See http://math.stackexchange.com/questions/1120480/convergence-of-sum-n-1-infty-1n-frac2n-sin-2nx-n/1120483#1120483 – lab bhattacharjee Jan 26 '15 at 16:06
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    Would it help to use $\cos^2\theta = \frac{1 + \cos 2\theta}{2}!$? – draks ... Jan 26 '15 at 16:07
  • Using @draks..'s hint you should be able to compute a closed form for your series :) – r9m Jan 26 '15 at 16:13
  • I'll give it a shot. Thanks – AlonAlon Jan 26 '15 at 16:21
  • @imranfat: this is intermediate between a straight and an alternating harmonic series. How can you conclude ? –  Jan 26 '15 at 16:21
  • @YvesDaoust I am just trying to understand that if the alternating harmonic series is convergent, how the addition of a squared cosine term in the numerator, that only produces values between zero and 1, could change the series into something divergent. – imranfat Jan 26 '15 at 16:47
  • @imranfat: if the squared cosine produced $(0,1,0,1,0,1,\cdots)$, the sum would diverge because there would be no cancellation anymore. –  Jan 26 '15 at 16:54
  • In this case that would never happen, but I do get to see why my reasoning is weak in terms of showing that the series is convergent... – imranfat Jan 26 '15 at 17:58

1 Answers1

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First write

$$\sum_{n = 1}^\infty \frac{(-1)^n\cos^2 n}{n} = \sum_{n = 1}^\infty \left(\frac{(-1)^n}{2n} + \frac{(-e^{2i})^n}{4n} + \frac{(-e^{-2i})^n}{4n}\right)$$

using the identities $$\cos^2 x = \frac{(e^{ix} + e^{-ix})^2}{4} = \frac{1}{2} + \frac{e^{2ix}}{4} + \frac{e^{-2ix}}{4}.$$ The series $\sum_{n = 1}^\infty (-1)^n/(2n)$ converges by the alternating series test, and series $\sum_{n = 1}^\infty (-e^{2i})^n/(4n)$ and $\sum_{n = 1}^\infty (-e^{-2i})^n/(4n)$ converge by Dirichlet's test. This shows that $\sum_{n = 1}^\infty (-1)^n\cos^2(n)/n$ converges. It's value can be computed:

\begin{align}&\sum_{n = 1}^\infty \frac{(-1)^n}{2n} + \sum_{n = 1}^\infty \frac{(-e^{2i})^n}{4n} + \sum_{n = 1}^\infty \frac{(-e^{-2i})^n}{4n}\\ & = -\frac{\log 4 + \log(1 + e^{2i}) + \log(1 + e^{-2i})}{4}\\ &= -\frac{\log 4 + \log[(1 + \cos 2)^2 + \sin^2 2]}{4}\\ &= -\frac{\log 4 + \log|2 + 2\cos 2|}{4}\\ &= -\frac{\log 4 + \log 2 + \log(1 + \cos 2)}{4}\\ &= -\frac{\log[8(1 + \cos 2)]}{4}. \end{align}

Note. In general, series of the form $$\sum_{n = 1}^\infty \frac{e^{in\theta}}{n},\quad \theta\in \Bbb R\setminus 2\pi \Bbb Z$$ are convergent, which I'll show by Dirichlet's test. Let $\theta \in \Bbb R\setminus 2\pi \Bbb Z$. The sequence $\left\langle \frac{1}{n}\right\rangle_{n = 1}^\infty$ decreases to $0$, and for every positive integer $N$,

$$\left\lvert \sum_{n = 1}^N e^{in\theta}\right\rvert = \left\lvert e^{i\theta}\frac{1 - e^{iN\theta}}{1 - e^{i\theta}}\right\rvert = \dfrac{\left\lvert\sin \frac{N\theta}{2}\right\rvert}{\left\lvert\sin \frac{\theta}{2}\right\rvert} \le \frac{1}{\left\lvert \sin \frac{\theta}{2}\right\rvert}.$$ Hence, by Dirichlet's test, the series $\sum_{n = 1}^\infty e^{in\theta}/n$ converges.

kobe
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