summation n=0 to infinity r^ncosn(theta)=rcos(theta)-r^2/1-2rcos(theta)+r^2
can neone prove this??
This is true iff $\displaystyle |r|< 1$, so that then we have the sum of a geometric sequence with quotient less than 1 in absolute value: $\displaystyle r^ne^{n i\theta}=r^n(\cos n\theta + i\sin n\theta), \mbox{using DeMoivre's theorem, so}$
$\displaystyle \sum\limits_{n=0}^\infty(re^{i\theta})^n=\frac{1}{ 1-re^{i\theta}}=\frac{1-re^{-i\theta}}{r^2+1-2r\cos \theta}$ <--correction done here
Now just separate real and imaginary parts above.
Tonio