Prove that $\frac{301}{900} $ number is $\sup$ of this set

Question

Problem Prove that $\frac{301}{900} $ is $\sup$ of this set:

I have the following set,

$$S=\{ 0.3 , 0.33 , 0.334 , 0.3344, 0.33444,... \}$$

I don't know how to do it, I know that I can represent $x=0.3344444...$ as an infnite series as: $$x=3\cdot 10^{-1}+3\cdot 10^{-2}+4\cdot 10^{-3}+4\cdot 10^{-5}+...$$ $$x=3\cdot 10^{-1}+3\cdot 10^{-2}+4 \sum_{n=3}^{\infty} (\frac{1}{10})^n$$

This will eventually converge to $\frac{301}{900} $, bu how can I prove that is the $\sup(S)$.

Answer 1

Hint: the sequence is increasing, so limit and sup coincide...

Answer 2

The theorem is called "théorème de la limite monotone" in French, but I can't find the English name.

In your case, for al $x\in S$, $l=\frac{301}{900}>x$

for all $\epsilon >0$, there is $k$ integer such as : $$\epsilon > 4\sum^{\infty}_{n=k}\frac{1}{10^n}=\frac{4}{10^k}\sum^{\infty}_{n=0}\frac{1}{10^n} =\frac{4}{10^k}\frac{1}{9}$$

Then $$x=\frac{33}{100}+ \sum^{k}_{n=3}\frac{1}{10^n}>l-4\sum^{\infty}_{n=k}\frac{1}{10^n}=\frac{33}{100}+ \sum^{k-1}_{n=3}\frac{1}{10^n}>l-\epsilon$$

So $l=\sup S$