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I have to establish the convergence of the following series: $$\sum_{n \ge1}\left(2+\sin n\right)^n\left(1-\dfrac{2}{n}\right)^{n^2}.$$

Using the root test I obviously obtained: $$\lim_{n\to\infty}\left(2+\sin n\right)\left(1-\dfrac{2}{n}\right)^{n}.$$ Does this limit even exist? I know that the limit of the second sequence is $e^{-2}$, but the sequence which contains $\sin n$ doesn't have a limit. Also, taken as a whole I don't know how to compute the limit. If it is indeed the case that it doesn't have a limit, then the root test is inconclusive so what test could I apply? Thank you for help!

Robert Z
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asd11
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1 Answers1

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Hint. For the root test, you are supposed to find $$\limsup_{n\to\infty}\left(\left(2+\sin n\right)\left(1-\dfrac{2}{n}\right)^{n}\right).$$

Robert Z
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  • Oh, I see. So we get $3e^{-2}$ taking the sin as 1, right? – asd11 Nov 28 '17 at 15:07
  • Yes. You may also say that the $\limsup\leq 3/e^2<1$. $\limsup\sin n=1$ but it is not a trivial result. See https://math.stackexchange.com/questions/1276422/show-that-limsup-of-sin-n-is-1 – Robert Z Nov 28 '17 at 15:08
  • Yeah. So the initial series is convergent! Thanks a lot! – asd11 Nov 28 '17 at 15:09