Talk:Sum of Infinite Geometric Sequence

Perhaps I'm confused, but when we say:


 * If $\left|{x}\right| < 1$, then by Power of Number less than One $x^{N+1} \to 0$ as $N \to \infty$.


 * Hence $s_N \to \dfrac 1 {1 - x}$ as $N \to \infty$

Isn't the the limit of the nth term a necessary but not a sufficient condition for convergence? --GFauxPas 19:08, 8 February 2012 (EST)


 * We're not proving convergence. We already know what the value of $s_N$ is. At this point we are not finding the sum of a series by adding up the sum of terms (we've already done that). $s_N$ is an element of a sequence. --prime mover 02:43, 9 February 2012 (EST)\


 * Ah, thanks. --GFauxPas 07:03, 9 February 2012 (EST)