# How to prove the series $\sum_{n=1}^{\infty} \frac{\sin(\frac{1}n)}n$ converges?

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

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.

• You might be able to do something with Dirichlet's Test – Davis Yoshida Mar 5 '14 at 5:52
• 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. – Lutz Lehmann Mar 5 '14 at 5:55
• Sorry, I meant sin, not cos! – rosstex Mar 5 '14 at 5:56
• Just compare with $1/n^2$. – Mariano Suárez-Álvarez Mar 5 '14 at 5:56
• $|\sin(x)|\le \min(1,|x|)$. – Lutz Lehmann Mar 5 '14 at 5:57

The series is convergent:

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

• 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? – rosstex Mar 5 '14 at 6:08
• @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! – Mariano Suárez-Álvarez Mar 5 '14 at 6:09
• @MarianoSuárez-Alvarez Sorry, I'm just getting used to Latex! You guys are very helpful, I'm on it – rosstex Mar 5 '14 at 6:10
• I now see the error I made, thanks! – rosstex Mar 5 '14 at 6:21

Hint: when $x\sim 0$, then

$$\sin x \sim x.$$

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