We know there is a rule that geometric series converges when $|r|<1$ , and a formula to calculate the sum as $S_n=\frac{a}{1-r}$ or as $S_n=\frac{a(1-r^n)}{1-r}$.
My question is: when I take $r$ infinitely close to $1$ (as $0.999999...$), wouldn't it be smaller than 1 again. I mean doesn't the equation $$ \lim_{x\to 1^-}\left(\sum_{n=0}^\infty x^n\right) $$ diverges where $x$ ($r$) is still smaller than $1$?