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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...2011-03-27

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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 $p_n$.)2011-09-01
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    @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.2011-09-01
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    What's the meaning of $\lim\ \cos p_n$? Dont't you mean $\lim \sum \cos p_n$ instead of $\lim\ \cos p_n$?2012-02-15
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    @draks: If $\sum a_n$ converges, then $a_n \to 0$.2012-02-15
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    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 :)2012-02-15
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    @draks: _If_ $\sum \cos p_n$ converges _then_ $\cos p_n \to 0$, and _then_ twin prime conjecture is false. I am not sure I understand your question.2012-02-15
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    Since 1.$\cos p_n \to 0$ doesn't seem to be right, because it's oscillating, 2. the sum doesn't converge, and therefore 3. the TPC is not false. Right?2012-02-15
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    @draks: The question was: "Is $\sum \cos p_n$ convergent?". I was trying to provide evidence that it($\sum \cos p_n$) is divergent/oscillating. All I say in the answer is "If $\sum \cos p_n$ is convergent, then $\lim \cos p_n = 0$ and then TPC is false". The second paragraph tries to add more evidence that it ($\sum \cos p_n$) is in fact not convergent. Basically, the assumption that $\sum \cos p_n$ is convergent is likely to be false. This is kind of like a "proof" by contradiction. Was I clearer?2012-02-15
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    I think I got it. Thanks.2012-02-15