Talk:Sum of Infinite Geometric Sequence

Perhaps I'm confused, but when we say:


 * If $\left|{x}\right| < 1$, then by Power of a 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)