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$$\sum_{n=1}^{\infty} \frac{\sin(\frac{1}n)}n$$

Using the comparison test/limit comparison test? I have tried the comparison test and several attempts at the limit comparison test, but everything I try points to divergence, which I know isn't true.

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  • $\begingroup$ You might be able to do something with Dirichlet's Test $\endgroup$ – Davis Yoshida Mar 5 '14 at 5:52
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    $\begingroup$ Compare with one half of the harmonic series, i.e., consider those n where $\cos\tfrac1n>\tfrac12$. And please give the reasons that make you exclude divergence. $\endgroup$ – Lutz Lehmann Mar 5 '14 at 5:55
  • $\begingroup$ Sorry, I meant sin, not cos! $\endgroup$ – rosstex Mar 5 '14 at 5:56
  • $\begingroup$ Just compare with $1/n^2$. $\endgroup$ – Mariano Suárez-Álvarez Mar 5 '14 at 5:56
  • $\begingroup$ $|\sin(x)|\le \min(1,|x|)$. $\endgroup$ – Lutz Lehmann Mar 5 '14 at 5:57
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The series is convergent:

$$ 0\le \frac{\sin\frac 1n}{n} \le \frac{\frac 1n}{n} = \frac 1{n^2} $$ and $\sum \frac 1{n^2} <\infty$.

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  • $\begingroup$ Thanks! Would you mind pointing out the error in my logic of using the limit comparison test with $b_n=\frac1n$ and using L'Hôpital's? $\endgroup$ – rosstex Mar 5 '14 at 6:08
  • $\begingroup$ @rosstex, edti the question to include your reasoning. It is not possible to point out the error in what you did if you do not tell us, well, what you did! $\endgroup$ – Mariano Suárez-Álvarez Mar 5 '14 at 6:09
  • $\begingroup$ @MarianoSuárez-Alvarez Sorry, I'm just getting used to Latex! You guys are very helpful, I'm on it $\endgroup$ – rosstex Mar 5 '14 at 6:10
  • $\begingroup$ I now see the error I made, thanks! $\endgroup$ – rosstex Mar 5 '14 at 6:21
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Hint: when $x\sim 0$, then

$$ \sin x \sim x. $$

A related technique.

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    $\begingroup$ Thanks, I would upvote you if I had more reputation! $\endgroup$ – rosstex Mar 5 '14 at 6:22
  • $\begingroup$ You are welcome. $\endgroup$ – Mhenni Benghorbal Mar 5 '14 at 7:44

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