So I have proven the limit $$ \lim_{n \to +\infty}n^3 \left(\frac{1}{n}-\sin{\left( \frac{1}{n}\right)} \right) = \frac{1}{6}. $$ I then have to prove that the series $$\sum_{n=1}^{+\infty} \left(\frac{1}{n}-\sin{\left( \frac{1}{n}\right)} \right)$$ converges using the limit. I don't know how to go on from here and I'm really confused about using the limit for this task.
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1Do you know the Limit Comparison Test? – Sewer Keeper Jun 02 '20 at 13:37
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@SewerKeeper : what are we comparing this to? – jimjim Jun 02 '20 at 13:46
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It's not in our book, so I don't think we can use that. @SewerKeeper Thank you for editing the post! – Zev Love X Jun 02 '20 at 13:47
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1@Arjang to $\frac{1}{n^3}$, right? – Sewer Keeper Jun 02 '20 at 13:50
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@ZevLoveX which tool can you use? – Sewer Keeper Jun 02 '20 at 13:50
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Using the site that Sewer provided. -Ratio test -Root test -Integral test -Direct comparison test -Cauchy condensation test – Zev Love X Jun 02 '20 at 13:54
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Why the obsession with this particular series? – Angina Seng Jun 02 '20 at 13:55
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@AnginaSeng Part of an assignment – Zev Love X Jun 02 '20 at 13:57
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Others have asked about it recently.... – Angina Seng Jun 02 '20 at 14:05
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@SewerKeeper : yes, well spotted. I can't see how I would write it down clearly though. – jimjim Jun 02 '20 at 14:09
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@AnginaSeng Can you point me to it? – Zev Love X Jun 02 '20 at 14:17
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@ZevLoveX note that your series is a series of positive terms. Hence since the general term is asymptotic to $1/n^3$ by your limit, it follows that your series converges or diverges with the series $\sum 1/n^3$. Since this is convergent, yours also is. – dude Jun 02 '20 at 14:54
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Fix $\epsilon\in(0,\,1/6)$. Some $N\in\Bbb N$ satisfies $n^3(1/n-\sin(1/n))\in(1/6-\epsilon,\,1/6+\epsilon)$ for all integers $n\ge N$. Since $A:=\sum_{n=N}^\infty\frac{1}{n^3}$ is finite, $\sum_{n=N}^\infty(1/n-\sin(1/n))\in((1/6-\epsilon)A,\,(1/6+\epsilon)A)$. This sum of positive reals is therefore convergent.
J.G.
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