As part of alarger proof, I'm trying to show that if $x$ and $y$ agree up to the $(N+1)$st digit in their decimal expansions (e.g., both are $1.41412 \ldots$ or elements in the sequence of successive decimal approximations of $\sqrt{2}$), then $|x-y| \leq \frac{1}{10^N}$. I can't figure out why this is $\frac{1}{10^N}$ instead of, say, $\frac{1}{10^{N-1}}$ or $\frac{1}{10^{N+1}}$.
Here is my attempt at showing this. Write out the decimal expansions of $x$ and $y$ as follows: $$ x = \sum\limits_{i=0}^{\infty} a_i 10^{-i}, \; y = \sum\limits_{i=0}^{\infty} b_i 10^{-i}. $$ Without loss of generality, I'm going to assume that $x \geq y$, so $|x-y| = x-y$, so it suffices to consider $x-y$. We have $$ x - y = \sum\limits_{i=0}^{\infty} (a_i - b_i)10^{-i}. $$ For each $0 \leq i \leq N$, $a_i = b_i$, so $a_i - b_i = 0$. We drop these terms. $$ x - y = \sum\limits_{i=N+1}^{\infty} (a_i - b_i) 10^{-i}. $$ At this point, I'm stuck. One thing I can try is to rewrite this as a geometric series. I would need to in some way eliminate $a_i - b_i$, which I can't figure out how to do.
Using geometric is a right direction, notice that the digits are bounded between $0$ and $9$.
\begin{align} x-y &\le \sum_{i=N+1}^\infty (a_i-b_i)\cdot 10^{-i}\\ &\le \sum_{i=N+1}^\infty |a_i-b_i|\cdot 10^{-i}\\ &\le 9\sum_{i=N+1}^\infty 10^{-i}\\ &=9\cdot \frac{10^{-(N+1)}}{1-0.1}\\ &=10^{-N} \end{align}