Show that
$$ \sum_{n=1}^\infty {\sin n \over n} = {-1 + \pi \over 2}$$

Question

experimentx · Answer

anonymous · Answer

it has to do something with fourier series...or maybe there is an easier way

anonymous · Answer

$$ \sum_{n=1}^\infty {\sin n \over n} = {-1 + \pi \over 2}$$

experimentx · Answer

yep!!

experimentx · Answer

can't help myself with any other method ... few days ago i tried fooling around ... i couldn't do it.

anonymous · Answer

experimentx · Answer

experimentx · Answer

looks this yahoo is worth try !! i haven't seen that one.

anonymous · Answer

but there is hole in yahoo proof .... is this converges?

$$\sum_{n=1}^{\infty} e^{inz}$$ for $|z|<1$  and  $z=1$

anonymous · Answer

if converges for $|z|<1$ we must go to Abel's theorem for $z=1$

experimentx · Answer

wonskian =/= 0 ??

experimentx · Answer

this is |dw:1345820356768:dw|
this might converge (let's say it for now)
the other one 
e^(inx) diverges (or not known) ... i guess

experimentx · Answer

hold on let's do this
|dw:1345820533021:dw|