Property $d(x_n,x_{n+1})\le 10\cdot2^{-n}$ in a complete metric space 
Let $(X,d)$ be a complete metric space and $(x_n)$ a sequence in $X$ such that $$d(x_n,x_{n+1})\le 10\cdot2^{-n}, \text{ for all $n$.}$$ Show that the sequence converges to some $a \in X$. Show that $d(x_5,a) < 1.$

Since $X$ is a complete metric space every Cauchy sequence in $X$ will converge. If $(x_n)$ is Cauchy I have that $$d(x_n,x_m)< \varepsilon, \text{ for $n,m \ge K \in \mathbb{N}.$}$$
Now if I fix $m = n+1$, then $$d(x_n,x_{n+1}) < \varepsilon, \text{ for $n,n+1 \ge K$.}$$
And since it's Cauchy there must exist $a$ such that the sequence converges to it? I might be headed completely in the wrong direction here by assuming that $(x_n)$ would be Cauchy since that wasn't stated, but I'm not sure what other properties the completeness would give me? How should I approach from here if this is even in the right direction to begin with?
 A: You cannot fix $m=n+1$ at all. Just note that for $n < m$, by the triangle inequality over $m-n-1$ steps:
$d(x_n, x_m) \le \sum_{i=0}^{m-n-1} d(x_{n+i}, x_{n+i+1})\le 10\sum_{i=0}^\infty \frac{1}{2^{n+i}} = 20\frac{1}{2^n}$ so that $(x_n)_n$ is Cauchy and hence converges to some $a \in X$.
The above inequality $d(x_n, x_m) \le 2\frac{1}{2^n}$ does not depend on $m$, so letting $m \to \infty$ we get $d(x_5,a) \le \frac{20}{32} <1$.
A: To prove that the sequence $\{x_n\}$ is Cauchy, we need to show that for any given $\epsilon > 0$, $\exists N \in \mathbb{N}$ s. t.  $n, m \geq N  \implies d(x_n, x_m) < \epsilon$.
We have, $d(x_n,x_{n+1})<10.2^{-n}$
$\therefore$ with $m>n$ and by repeated application of triangle inequality,
$d(x_n,d_m) \leq d(x_n,x_{n+1}) + d(x_{n+1},x_{n+2}) + \ldots d(x_{m-1},x_{m})$
$< 10.2^{-n}+10.2^{-(n+1)}+\ldots+10.2^{-(m-1)}+10.2^{-m} = 10.\frac{2^{-n}(1-(1/2)^{n-m})}{1-1/2} < 10.\frac{2^{-n}}{1/2}=\frac{5}{2^{n-2}} < \epsilon$
Hence, given any $\epsilon > 0$, we can choose $N=\lceil2+\log_2{\frac{\epsilon}{5}}\rceil$, s.t.,   $n, m \geq N  \implies d(x_n, x_m) < \epsilon$
