Odd cosn problem

For n= 1, cos(pi/1)= -1.

For n> 1, n odd, essentially you are adding the real parts of the 2nth roots of unity. Since those roots are symmetric about the imaginary axis, the sum is 0.

 

Ahh that it was... almost racked my brains out 'cause "π" I read as n ( not [itex]\pi[/itex]) ...

 

I am not sure what is being asked. Is this [tex]\sum cos(n_i)/n_i, or \sum cos(pi*n_i/n_i), or what?[/tex]

 

I interpreted as sum of [itex] cos(i\pi/n)[/tex] for i= 1 to n-1.