2

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 $1$?

ArthD21
  • 382
Memduh Yılmaz
  • 121
  • 6
    No. $0.999999\dots = 1$. In the real number system, there's no such thing as "infinitely close" without being there. Indeed, $\lim\limits_{r\to 1^-} \frac1{1-r}$ does not exist. – Ted Shifrin Aug 09 '22 at 16:44
  • 2
    Welcome to Math.SE! <> As Ted says, two real numbers are "infinitely close" if and only if they are equal. Separately, the limit expression in the question is "$+\infty$" if we use the standard definition of a limit diverging to infinity. – Andrew D. Hwang Aug 09 '22 at 16:46
  • 1
    So, you mean that limits are not real numbers, and we can only use real numbers on series as r? – Memduh Yılmaz Aug 09 '22 at 17:56
  • Regarding the prior comment, (i) not exactly; this particular limit is not a real number, but $+\infty$. (As a colleague put it, this is a very particular way of saying "the limit does not exist.") (ii) If it matters, the geometric series makes sense for certain types of objects other than real numbers. Here, we can meaningfully restrict ourselves to real numbers $x$ satisfying $-1 < x < 1$. (But again, as Ted notes, $0.99\overline{9} = 1$ does not satisfy the necessary strict inequality.) – Andrew D. Hwang Aug 09 '22 at 20:25
  • Perhaps a little more concretely, if we use geometric ratios $ \ r_n \ = \ 1 - 10^{-n} \ \ , $ then allowing the ratios to "approach 1 from below" would produce infinite geometric sums $$ \lim_{n \ \rightarrow \ \infty} \ S_n \ \ = \ \ \lim_{n \ \rightarrow \ \infty} \ \frac{a}{1 \ - \ (1 \ - \ 10^{-n})} \ \ = \ \ \lim_{n \ \rightarrow \ \infty} \ 10^n · \ a \ \ ,$$ which is plainly a limit that does not exist. –  Oct 06 '22 at 20:49

0 Answers0