Series

**Nacho** · October 15th 2008, 04:57 PM

Converge?

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

thanks

**ThePerfectHacker** · October 15th 2008, 05:52 PM

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.

**ThePerfectHacker** · October 15th 2008, 06:21 PM

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.

**Nacho** · October 15th 2008, 06:40 PM

Thanks!

but I don`t know about complex number

and haven`t demostration of Dirichlet`s test

again, than you ThePerfectHacker

**ThePerfectHacker** · October 15th 2008, 08:43 PM

Originally Posted by Nacho

Thanks!

but I don`t know about complex number

and haven`t demostration of Dirichlet`s test

again, than you ThePerfectHacker

How can I help you?

What do you need explained?

**Nacho** · October 16th 2008, 03:59 PM

Originally Posted by ThePerfectHacker

How can I help you?

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

**Krizalid** · October 17th 2008, 11:00 AM

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

**ThePerfectHacker** · October 17th 2008, 11:19 AM

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.

**Krizalid** · October 17th 2008, 11:33 AM

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

Originally Posted by ThePerfectHacker

But my method should work if fixed.

Yes.

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