TITLE: Showing limit of sequence $\left(\frac{3}{10}, \frac{33}{100}, \frac{333}{1000}, \dots\right)$
QUESTION [6 upvotes]: I'm trying to calculate the limit of the following sequence:
$$
(s_n) = \left(\frac{3}{10}, \frac{33}{100}, \frac{333}{1000}, \dots\right).
$$
Clearly, $(s_n) \to 1/3$, but I'm not sure how to show it rigorously. I need to be able to generate this to cases where $3$ is any integer $a \in \{ 1, 2, \dots, 9 \}$. I realize that the general limit would be $a/9$, but showing it is proving to be difficult. I'm looking for a way to express the numerator as a function of $n$, where $n$ is the "length" of the number. That is, if $a=4$ then $44$ corresponds to $n=2$ and $444$ corresponds to $n=3$. Any ideas?

REPLY [4 votes]: HINT :
$$
s_n=3\sum_{k=1}^n10^{-k}
$$
It's clearly the summation yields geometry progression with $r=\dfrac1{10}$.