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Consider the series

$$\sum_{n=0}^{\infty}\cos^n(n)$$

I think that the root test is inconclusive, because

$$\limsup_n \sqrt[n]{|\cos^n(n)|}=\limsup_n|\cos(n)|\leq 1$$

once we can approximate $\pi$ by rational numbers, there will always be some $i$ and $j\in\mathbb{N}$ such that $|j\pi-i|<\varepsilon$, for every $\varepsilon>0$ that we choose. And in this case $|\cos(i)-1|<\delta$.

Nevertheless, it seems that it converges. I can't think of any other convergent series to compare with it.

My question is: how can I prove that this series converges?

Edit: Actually, this series diverges, as you can see in tmyklebu's answer. I made a fortran program and here are some values of the sequence of the partial sums:

n     S_n
10    1.5898364866640549
100   7.8365722183614510
1000  24.825953005207236
10000 79.232008037801393
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  • $\begingroup$ Actually that $\limsup =1.$ $\endgroup$
    – zhw.
    Feb 28, 2016 at 0:15
  • $\begingroup$ I would consider the fact that $\cos(n)$ for $n \in \mathbb{Z}$ is always less than 1, since you have an integer argument. Of course the only time it equals one is when you have an integral multiple of $\pi$. However, you would approximate $\pi$ arbitrarily closely an infinite amount of times, but you would have $(1-\epsilon)$ to an aribtrarily large power, which still tends to 0. This isn't rigorous of course, but may give some intuition behind the convergence. $\endgroup$
    – Rellek
    Feb 28, 2016 at 0:15
  • $\begingroup$ @Rellek I thought of this idea too. It is a pity that it doesn't work. $\endgroup$
    – Larara
    Feb 28, 2016 at 0:47

1 Answer 1

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It doesn't. That series has a lot of terms near $\pm 1$; in particular, I can show it has infinitely many terms whose absolute value is larger than $\frac12$:

The inequality $|\pi - p/q| < 1/q^2$ is satisfied by infinitely many pairs of positive integers $(p,q)$. (This is Dirichlet's approximation theorem.) Let $(p,q)$ be one such pair with $q > 8$. Then, for some real $r$ with $|r| < 1/q$, we have $$|\cos(p)| = |\cos(q \pi + r)| \geq 1 - r^2 \geq 1 - 1/q^2.$$ Then $$|\cos(p)^p| \geq (1 - 1/q^2)^p \geq 1 - p/q^2 \geq 1 - 4/q > 1/2.$$

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    $\begingroup$ Indeed, the numerators of the convergents to $\pi$, which can be found from its continued fraction, furnish examples of such numbers $p$. The first twenty numerators are 3, 22, 333, 355, 103993, 104348, 208341, 312689, 833719, 1146408, 4272943, 5419351, 80143857, 165707065, 245850922, 411557987, 1068966896, 2549491779, 6167950454, 14885392687; and we can see computationally that $|\cos(p)^p|$ rapidly approaches $1$ for this sequence. $\endgroup$
    – Greg Martin
    Feb 28, 2016 at 0:23
  • $\begingroup$ @GregMartin ... So Larara used only 10000 terms, and that wasn't enough. $\endgroup$
    – GEdgar
    Feb 28, 2016 at 0:49
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    $\begingroup$ This conclusion is pretty obvious if we know that $\cos[\mathbb{N}]$ is actually dense over $(-1,1)$. $\endgroup$
    – Rasmus Erlemann
    Feb 28, 2016 at 0:49
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    $\begingroup$ @REr but what about $\lim_{n\to\infty}\cos^n[\mathbb{N}]$? $\endgroup$
    – Larara
    Feb 28, 2016 at 0:53
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    $\begingroup$ @REr Not quite. We need to know the existence of a subsequence which approaches 1 fast enough to overcome the raising to the $n$th power. Merely knowing that $\cos[\mathbb{N}]$ is dense is not enough. $\endgroup$
    – Eric M. Schmidt
    Feb 28, 2016 at 0:54

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