Every infinite sequence $ \ 0. a_1a_2a_3 \cdots \cdots $ represents a real number $ \ x \in [0,1] $ and $ \ \sum\limits_{i=1}^{n} \frac{a_i}{10^i} \leq x \leq \sum\limits_{i=1}^{n} \frac{a_i}{10^i} +\frac{1}{10^n} $ in decimal system for all $ n \in \mathbb{N}$.
Prove the similar inequality $ \ \sum\limits_{i=1}^{n} \frac{a_i}{b^i} \leq x \leq \sum\limits_{i=1}^{\infty} \frac{a_i}{b^i}+\frac{1}{b^n} \ $ for any base $b \neq 10 $.
Answer:
For any base $ \ b $ other than $ 10$ , a real number $ \ x \in [0,1] $ can be expressed as
$ x= \sum\limits_{n=0}^{\infty} a_n b^{-n} =a_0+\frac{a_1}{b}+\frac{a_2}{b^2}+\cdots +\frac{a_n}{b^n}+\cdots \cdots , \ \text{where $a_0 < 1$}. $
Thus, $ \ x \in [0,1] $.
But how to show that $ \sum\limits_{i=1}^{n} \frac{a_i}{b^i} \leq x \leq \sum\limits_{i=1}^{n} \frac{a_i}{b^i} +\frac{1}{b^n} $ holds good?
Help me showing this
Take the difference $$0\leq x - \sum\limits_{i=1}^{n} \frac{a_i}{b^i} = \sum\limits_{i=n+1}^{\infty} \frac{a_i}{b^i} \leq ...$$ and because $0\leq a_i \leq b-1$ and assuming an integer base $b>1$, we have $$ ... \leq \sum\limits_{i=n+1}^{\infty} \frac{b-1}{b^i}= \frac{b-1}{b^{n+1}} \left(\color{red}{\sum\limits_{i=0}^{\infty} \frac{1}{b^{i}}}\right)= \frac{b-1}{b^{n+1}} \left(\frac{1}{1-\frac{1}{b}}\right)=\frac{1}{b^n}$$ the sum in red is an infinite geometric progression with ratio $\left|\frac{1}{b}\right|<1$.