Prove that $d(x,y)=\sum_{i=1}^\infty \frac{|x_i - y_i|}{2^i}$ converges Question: Let S be the set of sequences of $0$s and $1$s. For $x = (x_1, x_2, x_3, ...)$ and $y = (y_1, y_2, y_3, ...)$. Define

$d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}$ 

Proof the infinite sum in the definition of $d(x,y)$ converges for all $x$ and $y$. 

Incomplete answer: Since the max of $d(x,y)$ happens when all elements of one of $x$ and $y$ is $1$ and the other is $0$, and the min of $d(x,y)$ happens when $x_n=y_n$ for all indices $n$, then 

$0=\sum_{i=1}^\infty \dfrac{0}{2^i}\leq d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}\leq \sum_{i=1}^\infty \dfrac{1}{2^i}=1 $

But how to prove that $d(x,y)$ converges to some point when $x$ and $y$ have fixed arbitrary elements?

Thank you.
The sequence $a_n = \sum_{k=1}^n \frac{|x_k-y_k|}{2^k}$ is monotonic and bounded from above, thus converges.

$$a_n \le \sum_{k=1}^n \frac{1}{2^k} \le  \sum_{k=1}^\infty \frac{1}{2^k} = 1$$
You have the right idea of bounding the the infinite sum. Now consider what the bounds would be if we started the series at the $n^{th}$ term. We can get that

$\sum_{i=n}^\infty \dfrac{|x_i-y_i|}{2^i}\leq \dfrac{1}{2^{n-1}}\to0$

as $n \to\infty$, which shows that the series converges.