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Show that the following series diverges $$\sum^{\infty}_{n=1} \sin{\left(\frac{1}{n}\right)}.$$

We do this by the comparison test. Let $a_{n} = \sin{\left(\frac{1}{n}\right)}$ Now the only test I can really apply here in my opinion is the comparison test however i'm not really sure what to compare this to. I want to find some $b_{n}<a_{n}$ and show that this $b_{n}$ diverges.

user2850514
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1 Answers1

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Using $\sin{x}\ge x-\frac{x^{3}}{3!}$. Thus $$\sin{\frac{1}{n}}\ge \frac{1}{n}-\frac{1}{6n^{3}}.$$

Let $b_{n}=1/n-1/(6n^{3})$. Note $\sum b_{n}$ diverges and $b_{n}\le a_{n}$ thus $\sum a_{n}$ diverges by the comparison test.

user2850514
  • 3,689
  • Why is the first inequality true? Not only using the Taylor series for $;\sin x;$ here could be seen as an overkill, but how would this help you to prove the inequality? – Timbuc Dec 28 '14 at 22:34