# Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy

Suppose the terms of the sequence of real numbers $\{a_n\}$ satisfy $|a_{n+1} - a_n| \le 2^{-n}$ for all $n$. Prove that $\{a_n\}$ is Cauchy.

My Work

So by the definition of a Cauchy sequence, for all $\varepsilon > 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. However, questions like this one make me understand that the $2^{-n}$ condition is necessary for this to be a true statement.

So I am wondering how to appeal to the Cauchy definition for this proof. Do I prove that every convergent sequence is therefore Cauchy, and then try to prove convergence?

As you said, you want to show that for any $$\epsilon>0$$ there is some $$n_0\in\Bbb N$$ such that $$|a_m - a_n|<\epsilon$$ whenever $$m, n \ge n_0$$. The trick is to figure out how big an $$n_0$$ you’re going to need to make sure that $$|a_m-a_m|<\epsilon$$ no matter how far apart $$m$$ and $$n$$ are, as long as they’re both at least $$n_0$$. Okay, suppose that we look at $$|a_m-a_n|$$ when $$m$$ and $$n$$ are not necessarily consecutive. There’s no harm in assuming that $$m\le n$$; then $$k=n-m\ge 0$$, and we’re looking at $$|a_m-a_{m+k}|$$. We only have a handle on the size of this number when $$k=1$$: if $$k=1$$, $$|a_m-a_{m+k}|\le 2^{-m}$$. But we also have the triangle inequality:

\begin{align*} |a_m-a_{m+k}|&=|(a_m-a_{m+1})+(a_{m+1}-a_{m+2})+\ldots+(a_{m+k-1}-a_{m+k})|\\ &\le|a_m-a_{m+1}|+|a_{m+1}-a_{m+2}|+\ldots+|a_{m+k-1}-a_{m+k}|\\ &<2^{-m}+2^{-(m+1)}+\ldots+2^{-(m+k-1)}\\ &<\sum_{k\ge m}\frac1{2^k}\\ &=\frac{\frac1{2^m}}{1-\frac12}\\ &=\frac1{2^{m-1}}\;. \end{align*}

Thus, if $$m,n\ge n_0$$, we automatically have $$|a_m-a_n|<\dfrac1{2^{m-1}}\le\dfrac1{2^{n_0-1}}$$. If we choose $$n_0$$ big enough so that $$\dfrac1{2^{n_0-1}}\le\epsilon$$, we’ll be in business. Is this always possible? Sure: just make sure that $$2^{n_0-1}\ge\dfrac1\epsilon$$, i.e., that $$n_0\ge\log_2\dfrac2\epsilon$$; this is certainly always possible.

Hint: Suppose $m>n$, then $$|a_{m}-a_{n}| = |a_{m}-a_{m-1}+a_{m-1}-a_{m-2}+a_{m-2}-a_{m-3}+\cdots+a_{n+1}-a_n|$$ $$\leq |a_{m}-a_{m-1}|+|a_{m-1}-a_{m-2}|+|a_{m-2}-a_{m-3}|+\cdots+|a_{n+1}-a_n|$$ $$\leq \sum_{i=n}^{m-1}2^{-n}$$

Can you see how to proceed?

• Thanks! I see the convergent series, which gives me the $N$ I need to pick. Aug 15, 2012 at 14:24
• Does anyone know of a way to do this without using series? Sep 27, 2014 at 17:56

We just need $|x_{n+1}-x_n|\leq a_n$ where $a_n$ is a sequence such that $\sum_na_n<+\infty$. Indeed, for $m,n\in\Bbb N$, and denoting $S_N:=\sum_{j=1}^Na_j$, we have $$|x_{m+n}-x_n|\leq \left|\sum_{j=m}^{m+n-1}(x_{j+1}-x_j)\right|\leq \sum_{j=m}^{m+n-1}\left|x_{j+1}-x_j\right|\leq |s_{m+n-1}-s_{m-1}|,$$ and we conclude using the fact that $\{S_n\}$ is Cauchy. Actually, the condition is necessary.