I'm proving that the limit of the following recursive sequence is $\dfrac{10}{9}$:
$$s_0=1,\,s_n=s_{n-1}+\frac{1}{10^n}\quad\text{for }n\ge1$$
Showing that the sequence is monotonic was easy enough, but I admit I'm having more troubled with the boundedness.
I know I want to show $s_k\le\dfrac{10}{9}$ for all $k$, but I'm not sure if this is a proof by induction, or if I can use some simpler reasoning?
What I've tried:
$$\begin{align*}s_0=1&<\frac{10}{9}&\text{base case}\\\\
s_k&<\frac{10}{9}&\text{assumption}\\\\
s_{k+1}&=s_k+\frac{1}{10^k}\\
&<\frac{10}{9}+\frac{1}{10^k}\end{align*}$$
Somehow I need to show that $s_{k+1}<\dfrac{10}{9}$, but I feel like this isn't right.
Edit Let's say I'm writing a lesson plan for someone that hasn't learned about series yet. Is there no way to prove boundedness without introducing the concept of a series and writing the sequence as a geometric sum? The hints are great, just a bit beyond my scope.
Best Answer
This is a geometric series.
Since $$(1+x+x^2+\cdot+x^n)(1-x)=1-x^{n+1}$$
then $$s_n=\frac{1-\frac{1}{10^{n+1}}}{1-\frac1{10}}$$
and $\lim s_n=\frac{10}9$.
Also, this is $$1.1111\ldots11$$ which is the ninth part of $$9.999\ldots99<10$$