# Series

• Oct 15th 2008, 05:57 PM
Nacho
Series
Converge?

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

thanks
• Oct 15th 2008, 06:52 PM
ThePerfectHacker
Quote:

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.
• Oct 15th 2008, 07:21 PM
ThePerfectHacker
Quote:

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. (Thinking)
• Oct 15th 2008, 07:40 PM
Nacho
Thanks!

but I dont know about complex number (Worried) and havent demostration of Dirichlets test (Crying)

again, than you ThePerfectHacker :)
• Oct 15th 2008, 09:43 PM
ThePerfectHacker
Quote:

Originally Posted by Nacho
Thanks!

but I dont know about complex number (Worried) and havent demostration of Dirichlets test (Crying)

again, than you ThePerfectHacker :)

What do you need explained?
• Oct 16th 2008, 04:59 PM
Nacho
Quote:

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 :)
• Oct 17th 2008, 12:00 PM
Krizalid
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 $\frac{\pi-1}{2}.$
• Oct 17th 2008, 12:19 PM
ThePerfectHacker
Quote:

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.
• Oct 17th 2008, 12:33 PM
Krizalid
$\frac{1}{2i}\ln \frac{1-e^{-i}}{1-e^{i}}=\frac{1}{2i}\ln \left( -e^{-i} \right)=\frac{1}{2i}\ln \left( e^{(\pi -1)i} \right)=\frac{\pi -1}{2}.$

Quote:

Originally Posted by ThePerfectHacker

But my method should work if fixed.

Yes. (Sun)