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I dont know how to solve this problem, please help! Thank you!

$$\sum_{n=1}^\infty \cos (n^3-n)$$

amWhy
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Cj Cortez
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    For an infinite series to (possibly) converge, it is necessary that the limit of the summand as $n\to \infty$ is equal to $0$. Does that happen here? – projectilemotion Feb 09 '20 at 13:53

1 Answers1

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It diverges. For a series $\sum_{n \in \mathbb N} a_n$, it is a necessary condition for convergence that $a_n$ is a zero sequence. Since

$$\lim_{n \to \infty} \cos(n^3 - n)$$

doesn't exist because

$$\limsup_{n \to \infty} \cos(n^3-n) = \limsup_{n \to \infty}\cos n = 1 \neq -1 = \liminf_{n \to \infty} \cos(n^3-n),$$

the series $\sum_{n=1}^\infty \cos(n^3-n)$ diverges.

ATW
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    The fact that the limit doesn't exist needs proof. – GEdgar Feb 09 '20 at 14:44
  • Thanks for the comment, I edited my answer. – ATW Feb 09 '20 at 14:56
  • still needs work – Gabriel Romon Feb 09 '20 at 14:58
  • Unfortunately that $\limsup$ answer is incomplete. The assertion $\limsup_{n \to \infty} \cos(n^3-n) = \limsup_{n \to \infty}\cos n$ needs proof. But do not give up: of course this is the correct way to show what the OP wants. – GEdgar Feb 09 '20 at 14:58
  • This is the kind of problem where the answer is super obvious, but where a proof is not completely straight forward. See for example https://math.stackexchange.com/questions/509678/how-prove-this-lim-n-to-infty-sinnm-divergent ; https://math.stackexchange.com/questions/2373322/sin-n2-diverges for possible approaches – Winther Feb 09 '20 at 15:09
  • Hi! i still don't understand how it diverges, sorry. is there any possible test or more straightforward proof to show that the series diverges? – Cj Cortez Feb 10 '20 at 04:04