I think that the series $$\sum_{n=1}^{\infty} \dfrac{\sin(n)}{n}$$ converges conditionally, but I´m not able to prove it. Any suggestions ?
Asked
Active
Viewed 1,128 times
4
-
5Hint: Use Dirichlet's test. – Ayman Hourieh Jun 23 '13 at 01:11
-
1See this post to see why it doesn't converge absolutely. – David Mitra Jun 23 '13 at 01:19
-
You are right ! – Ezequiel Jun 23 '13 at 01:54
-
http://cazelais.disted.camosun.bc.ca/250/dirichlet.pdf – Ezequiel Jun 23 '13 at 02:09
-
2This is not a duplicate of the question in @DavidMitra's link. That question is a subset of this question, which requires one to show convergence of the sum without absolute values. – Cheerful Parsnip Jun 23 '13 at 02:57
-
Hello, welcome to Math.SE. I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – Lord_Farin Jun 23 '13 at 13:25
2 Answers
4
Yes, you are correct that your series converges:
So you need to use Dirichlet's Test, which you might be somewhat familiar with, in that the alternating series test is a particular case of the Dirichlet's Test. This includes a checklist of three conditions for a series like yours, which if satisfied, reveal the series is convergent. You just need to show that your series meets those conditions, and your conclusion follows.
Added: there is a corollary to the Dirichlet's test that tells that $$\sum_{n=1}^\infty a_n \sin n$$ converges whenever $\,\{a_n\}\,$ is a decreasing sequence that tends to zero. And indeed, in your case $$\sum_{n = 1}^\infty \frac{\sin n}{n},$$ we have that $a_n = \dfrac 1n$, which we know is a decreasing sequence which tends to zero as $n\to \infty$.
amWhy
- 209,954
-
1I wouldn't call Dirichlet's Test the "alternating series test". This last test is a particular case of Dirichlet's. – Pedro Jun 23 '13 at 02:30
-
Yes, @Peter, I was actually returning to the question to make that distinction. Thanks! – amWhy Jun 23 '13 at 02:34
-
Wait, no. Dirichlet's Test doesn't talk about conditional convergence, just convergence. Or are you saying something else in "This includes a checklist of three conditions for a series like yours, which if satisfied, reveal the series is conditionally convergent."? – Pedro Jun 23 '13 at 02:41
-
-
You did not answer the question. Is the convergence conditional? You need to show that $\sum_{n=1}^\infty \frac{|\sin(n)|}{n}=\infty$. – Mark Viola Sep 23 '20 at 23:26
1
Using Fourier series calculations it follows $$ \sum_{n=1}^{\infty}\frac{\sin(n x)}{n}=\frac{\pi-x}{2} $$ for every $x\in(-\pi,\pi)$. Your sum is $\frac{\pi-1}{2}$.
vesszabo
- 3,481