Let $(X,d)$ be a metric space and $\{a_n\in X\}_{n\in \mathbb{N}}$ a sequence in $X$. Is the following true?
$$d(a_{n+1},a_n)<2^{-n}, \forall n\in \mathbb{N} \implies \{a_n\}_{n\in \mathbb{N}} \text{ is a Cauchy sequence.}$$
All I could do is: Suppose that $m>n$. Using the triangle inequality we have
\begin{eqnarray} d(a_m,a_n)\leq\sum_{i=1}^{m-n}d(a_{m-i+1},a_{m-i})\leq \sum_{i=1}^{m-n}2^{m-i}=2^m\sum_{i=1}^{m-n}2^{-i}=2^m\dfrac{2^{n-m-1}-2^{-1}}{2^{-1}-1}=2^{m}-2^{n}, \end{eqnarray}
which seems not to help showing that the sequence is Cauchy.
I think this must be true because I am reading a proof in a book at which the autor seems to use this fact. He comes (inside his context) to $d(a_{n+1},a_n)<2^{-n}, \forall n\in \mathbb{N}$ and then says "in particular, $\{a_n\}_{n\in \mathbb{N}}$ is a Cauchy sequence", without any explanation.
Could this "fact" be false, even this way?