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Let the series $\displaystyle\sum_{n=1}^\infty a_n$ with $a_n=\cos(\pi \sqrt{n^2+n+1})$. I read that this series is convergent. I see that the series is divergent by the nth term divergence test since $a_n$ does not go to zero as $n$ goes to infinity. I don't see how to show that this series converge, thank you for your help!

palio
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1 Answers1

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Note that $$ \sqrt{n^2+n+1}=n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=n\left(1+\frac{1}{2n}+\frac{1}{2n^2}-\frac{1}{8n^2}+O(n^{-3})\right)\\=n+\frac{1}{2}+\frac{3}{8n}+O(n^{-2}), $$ so $$ \cos(\pi\sqrt{n^2+n+1})=\cos(\pi n + \frac{1}{2}\pi+\frac{3\pi}{8n}+O(n^{-2}))\\=(-1)^{n+1}\sin\left(\frac{3\pi}{8n}+O(n^{-2})\right) = (-1)^{n+1}\frac{3\pi}{8n}+O(n^{-2}). $$ The leading-order term, $(-1)^{n+1}\frac{3\pi}{8n}$, is an alternating series whose terms decrease in absolute value with a limit of zero, so it is (conditionally) convergent. The remainder term is absolutely convergent, since it is $O(n^{-2})$. Therefore, the series with the combined terms converges.

mjqxxxx
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    Nice (+1) beat me by a minute! It is fascinating how the subleading term enforces convergence due to the periodicity of sine. – tired Feb 22 '16 at 19:37
  • A series whose terms are asymptotic to those of a convergent alternating series does not always converge; e.g. consider what happens if we add 1/(n ln n) to the terms of the alternating harmonic series. The answer has been edited to repair this gap and then the editor gave +1 vote. – Oscar Lanzi Feb 24 '16 at 23:16
  • If your answer is independent. go ahead and submit it anyway. I do. – marty cohen Feb 24 '16 at 23:50
  • @OscarLanzi: Thank you, you're right that the size of the remainder term is also relevant. – mjqxxxx Feb 24 '16 at 23:51
  • @mjqxxxx: You're welcome. When using adymptotic analysis to prove that a series converges, calculate with sufficient accuracy so that the remainder-term series converges absolutely. Then you are never wrong. – Oscar Lanzi Feb 25 '16 at 00:15