Converge?

$\displaystyle

\sum\limits_{n = 1}^\infty {\frac{{\sin (n)}}

{n}}

$

thanks

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- Oct 15th 2008, 04:57 PMNachoSeries
Converge?

$\displaystyle

\sum\limits_{n = 1}^\infty {\frac{{\sin (n)}}

{n}}

$

thanks - Oct 15th 2008, 05:52 PMThePerfectHacker
Yes. It can be proven by Dirichlet's test that $\displaystyle \sum_{n=1}^{\infty}\frac{\sin n}{n}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n}$ converge.

- Oct 15th 2008, 06:21 PMThePerfectHacker
It is more interesting to sum this series.

Remember that $\displaystyle 2i\sin (x) = \sinh (ix)$.

Therefore,

$\displaystyle \sum_{n=1}^{\infty} \frac{\sin (n)}{n} = \frac{1}{2i} \left( \sum_{n=1}^{\infty} \frac{e^{in}}{n} - \sum_{n=1}^{\infty} \frac{e^{-in}}{n} \right)$

Now use the formula,

$\displaystyle \sum_{n=1}^{\infty} \frac{z^n}{n} = - \log (1-z)$ for $\displaystyle |z| < 1$

Therefore, we get,

$\displaystyle \frac{1}{2i}\left( \log (1 - e^{-i}) - \log (1 - e^{i}) \right) = \sin 1$

I hope I did not make any mistakes. (Thinking) - Oct 15th 2008, 06:40 PMNacho
Thanks!

but I don`t know about complex number (Worried) and haven`t demostration of Dirichlet`s test (Crying)

again, than you ThePerfectHacker :) - Oct 15th 2008, 08:43 PMThePerfectHacker
- Oct 16th 2008, 03:59 PMNacho
Maybe we can solve delimiting?

I always find other serie minor, but this coverge or a major, but this diverge...

Too I try whit quotien rule o the integral, but never obey the hypothesis :(

Moreover I am in first year of university, so I dont know powerful tool...

anyways...thank you ThePerfectHacker :) - Oct 17th 2008, 11:00 AMKrizalid
But you asked if the series does converge or not and that was answered. Now, wanna know how to find the value of that series? Its value is $\displaystyle \frac{\pi-1}{2}.$

- Oct 17th 2008, 11:19 AMThePerfectHacker
- Oct 17th 2008, 11:33 AMKrizalid