How do we prove that if $M$ is complete, any sequence $x_n \in M$ satisfying $d(x_n,x_{n+1}) < 2^{-n}$ converges in $M$? I'd like to show that if $M$ is a metric space $(M,d)$, and is complete, any sequence $x_n \in M$ satisfying $d(x_n,x_{n+1}) &lt; 2^{-n}$ converges in $M. 

My proof is the following:

I think that this question is the same as showing that if all Cauchy sequences, say  $x_n \in M$ are convergent in $M$ (by definition of complete), then that sequence $x_n$ must also satisfy the condition $d(x_n,x_{n+1}) &lt; 2^{-n}$. Is the proof as simple as choosing $\epsilon \leq 2^{-n}$ and then letting $m=n+1$? Since if we have that we can say:

Suppose $x_n \in M$ is a Cauchy sequence. Hence, it must be convergent. Then, by definition this means that given $\epsilon &gt;0$, there exists an $N$ such that if $m,n \geq N$, $d(x_m,x_n)&lt; \epsilon$.

Hence, just choose $\epsilon$ small enough so that $\epsilon \leq 2^{-n}$ and then let $m=n+1$. Then, pugging into the above, we will have the following statement:

given $0&lt;\epsilon  \leq 2^{-n}$, there exists an $N$ such that if $m=n+1,n \geq N$, $d(x_m,x_n) = d(x_{n+1}, x_n) &lt; \epsilon \leq 2^{-n}$. Hence, $x_n \in M$ satisfies

$d(x_{n+1}, x_n) &lt; 2^{-n}$. And, since $x_n$ is a Cauchy sequence in a complete space, its also convergent. 

I am wondering if the above makes sense. I am a bit wary of the part where I "choose" and "fix" the $\epsilon$. Thank you!
The proof as stated is wrong. A simple counterexample is $(1,0,\frac{1}{3},0,\frac{1}{5},0,\ldots)$. Moreover, it tries to prove the wrong implication. What you need to prove is the following: if $(x_n)$ is a sequence in $M$ such that $d(x_n,x_{n+1})&lt;2^{-n}$ for all $n\in \mathbb{N}$, then $(x_n)$ is a Cauchy sequence. This suffices, since $M$ is complete.

A proof for this statement would look like this: for all $m,n\in \mathbb{N}$ with $n\ge m$ it is true that $d(x_m,x_n)\le \sum_{k=m}^n d(x_k,x_{k+1})$. You can use induction to prove this. Choose $N$ such that $\sum_{k=m}^n 2^{-k}&lt;\epsilon$ for all $n,m\ge N$. You can do this, since the geometric series converges (absolutely). It follows that $d(x_m,x_n)\le \sum_{k=m}^n d(x_k,x_{k+1})\le \sum_{k=m}^n 2^{-k}&lt;\epsilon$ for all $m,n\ge N$, hence $(x_n)$ is a Cauchy sequence.
Your proof is correct. For some large enough $N$, the distance between subsequent terms $x_n,x_{n+1}$ in any Cauchy sequence is less then $2^{-n}$, for any $n&gt;N$.

However, this is not what you have to prove. You need the other implication.

Let $\{x_n\}^\infty$ be a sequence satisfying the property from your question. Let $\varepsilon&gt;0$. Now choose a natural $N$ such that $\varepsilon&gt;2^{-N}$. Such a number exists since the naturals are not bounded from above. Finally, consider any two natural numbers $n,m&gt;N$. Do you see, by the triangle inequality, that $d(x_n,x_m)&lt;2^{-N}&lt;\varepsilon$? This makes the sequence a Cauchy sequence, and therefore it converges by completeness.

Q.E.D.