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Let $p_n$ be the nth prime number, $p_1=2,p_2=3,p_3=5,\ldots$

How to prove this series converges/diverges?

$$\sum_{n=1}^\infty \cos{p_n}$$

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to converge, terms must go to zero, so the set of limit points of $p(n)\text{mod}2\pi$ must be contained in $\{\pi/2,3\pi/2\}$, which doesnt seem very likely... –  yoyo Mar 27 '11 at 19:06
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1 Answer

up vote 6 down vote accepted

If it converges, then this disproves the twin prime conjecture, I believe.

If $\lim\ \cos p_n = 0$ and the twin prime conjecture were true, then we would have that

as $p_n$ runs through the lower twin prime (i.e. both $p_n$ and $p_n + 2$ are primes),

$0 = \lim\ \cos (p_n + 2) = \lim\ (\cos p_n \cos 2 + \sin p_n \sin 2) = \pm \sin 2$

In fact,

If $\lim\ \cos p_n = 0$, then for any odd integer $M$, we must have that $\lim\ \cos (M\times p_n) = 0$ (as $\cos Mx$ can be written as an odd polynomial in $\cos x$), which I guess, implies that $ \lim\ \cos (2n+1) = 0$

If I remember correctly there was a previous question which disproved this.

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I don't follow your answer beyond the "In fact". You can conclude that $\cos(Mp_n) \to 0$ in the limit for any fixed odd integer $M$. But I don't think you can say much about numbers like $p_n^2$. (The point is that we still have a multiple of $p_n$, but it is not a constant times $p_n$.) –  Srivatsan Sep 1 '11 at 12:09
    
@Srivatsan: It was just an observation which might possibly used to prove that $\lim \cos(2n+1) = 0$. Note: "which I guess". So it is not a claim of a proof. Perhaps the "In fact" is misleading, I agree. –  Aryabhata Sep 1 '11 at 14:19
    
What's the meaning of $\lim\ \cos p_n$? Dont't you mean $\lim \sum \cos p_n$ instead of $\lim\ \cos p_n$? –  draks ... Feb 15 '12 at 20:08
    
@draks: If $\sum a_n$ converges, then $a_n \to 0$. –  Aryabhata Feb 15 '12 at 20:09
    
But how will $\cos p_n\to 0$? Is this how you showed, that the Twin Prime Conjecture is not disproved? (i like double negations :) –  draks ... Feb 15 '12 at 21:22
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