# Evaluate this limit $(\infty/\infty)$

I want to prove that $$1=0.999...$$ in this way, we have

$$0.9...9=0.9 \left(10^{-n-1}\right)\left(\sum_{i=0}^n 10^i\right)$$ Hence $$0.999...=\lim_{n \to \infty} 0.9 \left(10^{-n-1}\right)\left(\sum_{i=1}^n 10^i\right) \\ =0.9 \lim_{n \to \infty} \frac{\sum_{i=0}^n 10^i}{10^{n+1}}$$ Both the numerator and the denominator goes to $$\infty$$ so we xan use hôpital but even if we did we’d get $$\infty/\infty$$ again, how can I evaluate this limit?

• The series you you have is geometric. Use this fact! Dec 3, 2021 at 22:25
• Or use the discrete version of l'Hopital's rule, Stolz-Cesaro, because letting the upper bound of summation vary continuously is a little weird. Then your new limit is a ratio of the "last" term in the series over the corresponding power of ten. Then pretend you're using a continuous variable. Dec 3, 2021 at 22:27

When summing a geometric sequence, we want the common ratio to be less than $$1$$ in magnitude: $$\sum_{n=0}^\infty r^n = 1 + r + r^2 + \cdots = \frac{1}{1-r}$$ as long as $$\lvert r \rvert < 1$$.
Thus, to evaluate $$0.\overline{9} = 0.999\dots$$, we write it in terms of a geometric series with common ratio $$r=\tfrac{1}{10} < 1$$. Try it yourself, then click to reveal the calculation.
\begin{align} 0.\overline{9} &= \tfrac{1}{10} \bigl( 9 + \tfrac{9}{10} + \tfrac{9}{10^2} + \cdots \bigr) \\ &= \tfrac{9}{10} \bigl( 1 + \tfrac{1}{10} + \tfrac{1}{10^2} + \cdots \bigr) \\ &= \frac{9}{10} \sum_{n=0}^\infty \Bigl(\frac{1}{10} \Bigr)^n \\ &= \frac{9}{10} \cdot \frac{1}{1 - \tfrac{1}{10}} \\ &= \frac{9}{10 - 1} \\ &= 1 \end{align}