The point is that you still haven't told us precisely what the problem was!
The geometric sum \Sigma_{i=0}^\infty (cos(1))^n is a "geometric series", not a "Riemann Sum" (those are the finite sums used to define an integral). As Arildno said, you haven't stated whether that is 1 radian or 1 degree (although, I would say that, as long as we are talking about the cosine function as opposed to a trigonometry problem, radians should be assumed) and you haven't justified keeping only 4 significant figures.
Since the the geometric series \sigma_{i= 0}{\infty}ar^n has sum \frac{a}{1- r}, as long as |r|< 1, as is cos(1). In your problem, a= cos(1) (since the sum starts at k= 1, not 0) and r= cos(1) (not "(cos(1)k)" . The sum is \frac{cos(1)}{1- cos(1)}.
Since the common ratio is cos(1) rather than cos(1)k, you do not take the limit as k goes to infinity. Indeed, if that were correct, for |r|< 1, the sum of \sigma_{i=0}^{\infty}ar^n would always be just a.
