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$$\sum_{n=1}^\infty~\left|\frac{\cos2^n}{n}\right|$$ I just confused what to do. I am sorry, i don't understand. But this task is important for me. Can you give full solution?

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up vote 23 down vote accepted

Hint: Use the double-angle formula for $\cos x$ to show that in each pair of successive terms, at least one term has $|\cos 2^n| \geq \epsilon$ for some fixed universal $\epsilon > 0$. Then use that to prove the series diverges.

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I must use Cauchy's convergence test? – Simankov Apr 17 '14 at 21:36
@slmkarta Just take the terms in consecutive pairs, and show that for the $n$th pair you have that the sum of the two terms is greater than or equal to $\epsilon / 2n$. So then your series diverges by comparison test because $\sum_n (\epsilon/2)/n$ diverges. – user2566092 Apr 17 '14 at 21:43

I guess the homework tag has now become irrelevant and it's about understanding the argument. So let's spell out what user2566092 hinted.

The addition theorem for the cosine is

$$\cos (x+y) = \cos x \cos y - \sin x \sin y.$$

Specialising $y$ to $x$, we obtain the double-angle formula

$$\cos (2x) = \cos^2 x - \sin^2 x.$$

With the identity $\cos^2 x + \sin^2 x \equiv 1$, we can write that as

$$\cos (2x) = \cos^2 x - (1-\cos^2 x) = 2\cos^2 x - 1.$$

If $\lvert \cos x\rvert$ is small, then $\cos^2 x$ is very small, and $2\cos^2 x - 1$ is close to $-1$, and therefore $\lvert \cos (2x)\rvert$ is then not small.

We can quantify that: If $\lvert\cos x\rvert \leqslant \frac{1}{2}$, then $0 \leqslant \cos^2 x \leqslant \frac{1}{4}$, and $\cos (2x) \leqslant 2\cdot\frac{1}{4} - 1 = -\frac{1}{2}$, so then $\lvert \cos (2x)\rvert \geqslant \frac{1}{2}$. Thus in the sequence $a_n = \lvert\cos 2^n\rvert$, of every two consecutive terms, at least one is $\frac{1}{2}$ or greater. Therefore, grouping two consecutive terms, we see that

$$\sum_{n=1}^{2k} \left\lvert \frac{\cos 2^n}{n}\right\rvert \geqslant \sum_{m=1}^k \frac{\lvert\cos 2^{2m-1}\rvert + \lvert\cos 2^{2m}\rvert}{2m} \geqslant \sum_{m=1}^k \frac{1/2}{2m} = \frac{1}{4}\sum_{m=1}^k \frac{1}{m} > \frac{1}{4}\log k,$$

which shows that the series does not converge.

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