# $|x_{n + 1} - x_n| < \frac{1}{2^n} \Rightarrow (x_n)$ is Cauchy [duplicate]

Let $$(x_n)$$ be a real sequence with the property that for all $$n \in \mathbb{N}$$, $$|x_{n + 1} - x_n| < \frac{1}{2^n}$$ I want to show, using the definition of a Cauchy sequence, that $$(x_n)$$ must be Cauchy.

I have found that the property implies that for any $$(m, n) \in \mathbb{R}^2$$, assuming without loss of generality that $$m > n$$, it must be true that $$|x_n - x_m| \leq \sum\limits_{i = n}^m \frac{1}{2^i}$$

How can I proceed from there ? Is this even the right way to approach this problem?

## marked as duplicate by Martin R, Chinnapparaj R, Cameron Buie, mrtaurho, Adrian KeisterApr 12 at 13:49

• What is $k$, and how you got this inequality? Substituting $n = m + 1$ you get $x_n - x_m| < 0$ that can't be right. – mihaild Apr 12 at 11:22
• – Martin R Apr 12 at 11:26
• Fixed my question, thank you. I get this result by repeatedly applying the triangle inequality. I will go and check out this possible duplicate. – oranji Apr 12 at 11:27

• $$s_n = \sum_{k=1}^{n}(x_k - x_{k-1}) = x_n - x_0$$ is (absolutely) convergent
• $$\Rightarrow x_n = s_n + x_0$$ is convergent
• Btw, this only “hides” the triangle-inequality argument in the conclusion $\sum_{k=1}^{n}|x_k - x_{k-1}|$ convergent $\implies \sum_{k=1}^{n}(x_k - x_{k-1})$ convergent (and uses the Cauchy criterion, which is not needed to answer the original question). – Martin R Apr 12 at 11:46