# If $|x_{n+1} - x_n| < \epsilon,$ is $(x_n)$ a Cauchy sequence?

Suppose that $\, \forall \epsilon >0, \, \exists N, \,$ such that

$$|x_{n+1} - x_n| < \epsilon\, \quad \forall n \geq N.$$

Is the sequence $(x_n)$ a Cauchy sequence? If so, prove it; if not, give a counterexample.

My professor gave this counterexample: Let

$$x_n := \sqrt{n}$$

then $\, \forall\epsilon > 0\,$ take $\, N = \lfloor \frac{1}{\epsilon ^2}\rfloor +1 \,$ then we have

$$|x_n - x_{n+1} | = \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n} + \sqrt{n+1}} < \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} < \epsilon$$

But $(x_n)$ is divergent, so $(x_n)$ is not Cauchy.

I disagreed that this proof was valid, because the sequence $x_n := \sqrt{n} \,$ doesn't satisfy the hypothesis, particularly that if $\forall \epsilon > 0$

$$| x_{n+1} - x_n| < \epsilon\, , \quad \forall n \geq N$$ is satisfied, then the hypothesis requires that
$$| x_{n+2} - x_{n+1}| < \epsilon\, , \quad$$ is also satisfied, and so on.

Every single time I've disagreed with my analysis professor I've been wrong and I think this is not an exception. Really I would like someone to point out why the following proof (my attempt) that the sequence is Cauchy is wrong, please:

For $m \geq n$

\begin{align} |x_m - x_n| &= |x_m - x_{m-1} + x_{m-1} - \cdots + x_{n+2} - x_{n+1} - x_{n+1} + x_n| \\ &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+2} - x_{n+1}| + |x_{n+1} + x_n| \\ &< (m-n)\epsilon \end{align}

Therefore the sequence is Cauchy.

Where I think I'm wrong: the $(m-n)$ factor. I'm claiming that the error of each term can be made as small as we like so adding up however many terms makes no difference. Actually now, that I've typed this all out it seems obvious that I'm wrong, for example: I could have every term in a sum bounded by a one trillionth, but then add up a number of sequential terms well past one trillion, and make the total error as large as I would like, regardless of $\epsilon$. Can someone please confirm if that thinking is correct.

• $(m-n)$ might be much bigger than $\epsilon$ if they are allowed to be unrestrictedly distanced from each other. so much bigger in fact, that the product $(m-n)\epsilon$ doesn't go to zero.
– user123641
Dec 13, 2017 at 16:47
– user370967
Dec 13, 2017 at 18:11

I disagreed that this proof was valid, because the sequence $$x_n := \sqrt{n} \,$$ doesn't satisfy the hypothesis, particularly that if $$\forall \epsilon > 0$$

$$| x_{n+1} - x_n| < \epsilon\, , \quad \forall n \geq N$$ is satisfied, then the hypothesis requires that
$$| x_{n+2} - x_{n+1}| < \epsilon\, , \quad$$ is also satisfied, and so on.

Yes, and all the inequalities you list are satisfied. In particular, since

$$|\sqrt{n+1} - \sqrt{n}| = \frac{1}{\sqrt{n+1}+\sqrt{n}}\leq \frac{1}{\sqrt{n}}$$

you have, for every $$\epsilon > 0$$, the selection $$N=\lceil\frac{1}{\epsilon^2}\rceil$$ to prove it. From this alone, we can already be certain that your proof is incorrect (because $$x_n=\sqrt{n}$$ is obviously a good counterexample).

To see where your proof went wrong, you correctly prove that $$|x_m-x_n|<(m-n)\epsilon$$, however to prove that $$x_n$$ is cauchy, you should be able to prove that $$|x_m-x_n|$$ is arbitrarily small, but the expression contains the value $$(m-n)$$ which you were not able to bound. Yes, you could have every term in a sum bounded by one trillionth, but there is nothing to guarantee that $$m-n$$ is smaller than one quadrillion. The problem is you are given $$\epsilon$$ in advance, and then the inequality has to hold for every pair $$m,n$$.

In your proof $(m-n) \epsilon$ can't be made small as the indices are appearing in the sequence and as $m-n$ increases then the constant also changes.This can't be cauchy.

The argument given by the professor is right Note that $\frac{1}{\sqrt{n}} < \epsilon$ holds for all $n \ge N$ where $N= [1/\epsilon] +1$ So the condition $|x_{n+1}-x_n| < \epsilon$ holds for all $n \ge N$ where $N$ is defined above.

Your proof is wrong, because you 'real' epsilon is in fact

$${\rm epsilon}=\left(m-n\right)\varepsilon$$

and it depends on the indices of the sequence terms in question. Also, this epsilon can be arbitrary large if you choose $m=n+M$ with $M\gg\frac{1}{\varepsilon}$.