# Math Help - Series

1. ## Series

Converge?

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

thanks

2. Originally Posted by Nacho
Converge?

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

thanks
Yes. It can be proven by Dirichlet's test that $\sum_{n=1}^{\infty}\frac{\sin n}{n}$ and $\sum_{n=1}^{\infty} \frac{\cos n}{n}$ converge.

3. Originally Posted by Nacho
$
\sum\limits_{n = 1}^\infty {\frac{{\sin (n)}}
{n}}
$
It is more interesting to sum this series.

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

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

Now use the formula,
$\sum_{n=1}^{\infty} \frac{z^n}{n} = - \log (1-z)$ for $|z| < 1$

Therefore, we get,
$\frac{1}{2i}\left( \log (1 - e^{-i}) - \log (1 - e^{i}) \right) = \sin 1$

I hope I did not make any mistakes.

4. Thanks!

but I dont know about complex number and havent demostration of Dirichlets test

again, than you ThePerfectHacker

5. Originally Posted by Nacho
Thanks!

but I dont know about complex number and havent demostration of Dirichlets test

again, than you ThePerfectHacker
What do you need explained?

6. Originally Posted by ThePerfectHacker
What do you need explained?
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

7. 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 $\frac{\pi-1}{2}.$

8. Originally Posted by Krizalid
Now, wanna know how to find the value of that series? Its value is $\frac{\pi-1}{2}.$
Yes, thank you Krizalid. I thought I made a mistake. But my method should work if fixed.

9. $\frac{1}{2i}\ln \frac{1-e^{-i}}{1-e^{i}}=\frac{1}{2i}\ln \left( -e^{-i} \right)=\frac{1}{2i}\ln \left( e^{(\pi -1)i} \right)=\frac{\pi -1}{2}.$

Originally Posted by ThePerfectHacker

But my method should work if fixed.
Yes.