Proving completeness of a metric space Let $\mathbb{R}^{\infty}$ be the infinite Cartesian product of $\mathbb{R}$ with itself. Define the metric $d(x,y) = \sum_{t \in \mathbb{N} \cup \{0\}} \beta^{-t} \rho(x_t, y_t)$ where $\rho(x_t, y_t) = \min\{|x_t - y_t|, 1\}$ for some fixed $\beta>1$ and for all $x, y \in \mathbb{R}^{\infty}$ where $x = (x_0, x_1, \cdots)$ and $y=(y_0, y_1, \cdots)$. Prove that $(\mathbb{R}^{\infty}, d)$ is a complete metric space.
My attempt: I know that I need to show that every Cauchy sequence $\{x_n\} \subseteq \mathbb{R}^{\infty}$ converges to a limit $x \in \mathbb{R}^{\infty}$. So let $\{x_n\}$ be a Cauchy sequence contained in $\mathbb{R}^{\infty}$. Then, $\forall \varepsilon>0$ $\exists N$ such that $n,m > N$ implies $d(x_n, x_m) < \varepsilon$. 
Now $d(x_n, x_m) = \sum_{t \in \mathbb{N} \cup \{0\}}  \beta^{-t} \rho(x_{n, t}, x_{m, t})$. I need to somehow show that $\{x_n\}$ is actually converging and converges to a value in $\mathbb{R}^{\infty}$. How can I finish the proof?

You should describe a sequence $x$, and then show that $x_n \to x$.
A reasonable first guess is to define $x$ by setting $x_t = \lim_{n \to \infty} x_{n,t}$. It's not too hard to show that these $x_t$'s all exist. (Brief sketch: Let $\epsilon > 0$. For large enough $n$ and $m$, $d(x_n,x_m) < \epsilon \beta^{-t}$. We have $\beta^{-t} \rho(x_{n,t},x_{m,t}) \leq d(x_n,x_m)$. So, for large enough $n$ and $m$, $\rho(x_{n,t},x_{m,t}) < \epsilon$. This shows $(x_{n,t})_n$ is Cauchy.)
Next, showing $x_n \to x$, i.e., $d(x_n,x) \to 0$, is a little more involved but should be also possible. For $\epsilon > 0$, let $T$ be large enough so that the tail of the geometric sum $\beta^{-T} + \beta^{-T-1} + \dotsb < \epsilon$. Let $B = \beta^{-0}+\beta^{-1}+\dotsb+\beta^{-T+1}$. Let $N$ be large enough so that for $n \geq N$, $\rho(x_{n,t},x_t) < \epsilon/B$, for each $t=0,1,\dotsc,T-1$. (Why is this possible to say?) Now what is $d(x_n,x)$ for $n \geq N$? I'll leave the rest (and filling in details) to you.
