Prove or give a counterexample: if a sequence of real numbers $\{x_n\}$ from $n=1$ to $\infty$ has the property that for all $\epsilon >0$, there exists $N \in \mathbb N$ such that for all $n \ge N$ we have $|x_{n+1} - x_n| < \epsilon$, then $\{x_n\}$ is a convergent sequence. How is this different from the definition of a Cauchy sequence?

Attempt: For the second part of the problem, I know that it's different from the definition of a Cauchy sequence as it's taking the next part of the sequence and subtracting it from the current part of the sequence.

For the first part, I'm not sure how to go about doing this; originally, I thought I could do something such as $|x_{n+1} - x_n| < \epsilon/2$, in then use the triangle inequality. But I'm not sure we can do that. Thoughts and comments?

And apologies ahead of time for the lack of formatting.

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    $\begingroup$ Your condition is false. The qualifiers for Cauchy Convergence are very precise; to wit, the statement is that for any $\epsilon > 0$ I can find $N$ such that for any $n,m > N$ we have $|x_n - x_m| < \epsilon$. To see a concrete example, try $x_n = \sum_{i = 1}^n i^{-1}$. $\endgroup$ – A Blumenthal Mar 16 '13 at 23:57
  • $\begingroup$ Your idea with the triangle inequality will not work, as once $n$ is chosen, Cauchyness allows me to choose $m$ to be absurdly large, in which case I cannot get a good bound using only that $|x_n - x_{n+1}| <\epsilon$. $\endgroup$ – A Blumenthal Mar 16 '13 at 23:59
  • $\begingroup$ It might be worth noting that if $|x_{n+1}-x_n|<r^n$ for some $r<1$, the sequence is Cauchy. $\endgroup$ – Pedro Tamaroff Mar 17 '13 at 0:33


Try the sequence of harmonic numbers $\displaystyle\sum_{i=1}^n\frac{1}{i}$ and see if it converges.

  • $\begingroup$ Or (perhaps easier) $x_n = \sqrt{n}$. $\endgroup$ – GEdgar Mar 17 '13 at 0:19
  • $\begingroup$ @GEdgar That sequence does not have the required property. $\endgroup$ – Ian Coley Mar 17 '13 at 0:34
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    $\begingroup$ @FrankMcGovern Yes, it does. For each $n\in \Bbb N$, $$\sqrt {n+1}-\sqrt {n}=\frac{1}{2\sqrt \psi}$$ for some real $\psi\in[n,n+1]$. $\endgroup$ – Pedro Tamaroff Mar 17 '13 at 0:38
  • $\begingroup$ @Peter Oh, of course. I don't know what I was thinking. $\endgroup$ – Ian Coley Mar 17 '13 at 0:39
  • $\begingroup$ I tried the harmonic numbers for 1/i, and it does converge (to 0). Even if I were to use the example for my given question, (xn+1 − xn), the sequence would still converge, as it's still constrained by L+epsilon, L-epsilon. Maybe I'm not understanding this 100%. Could you elaborate some more, primarily on the example? I want to understand the example before we continue. $\endgroup$ – noname Mar 17 '13 at 2:37

Note that, a sequence ${x_n}$ is Cauchy if

$$ \lim_{n \to \infty} |x_{n+p}-x_n| = 0 \quad \forall p\geq 1. $$

Now, as an example of your case, try $x_n=\ln(n)$.



In general, every Cauchy sequence is $\mathbb{R}$ is convergent. (Three Steps)

  1. Prove that every Cauchy sequence is bounded.
  2. Use the Bolzano-Weierstrass Theorem to conclude that it must have a convergent subsequence.
  3. Show that a Cauchy sequence having a convergent subsequence must itself be convergent.

(The Bolzano-Weierstrass Theorem states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence.)

Note that the sequence given in your question is a special case of a Cauchy sequence.

To clarify the other assumptions:

(1) Harmonic series is NOT Cauchy.

Recall defintion: $\forall \varepsilon >0, \exists N \in\mathbb{N}$ such that $\forall m,n>N,|a_n-a_m|<\varepsilon $

Choose $m=2N$ and $n=N$. Then $$\left | \sum_{i=1}^{2n}\frac{1}{i} - \sum_{i=1}^{n}\frac{1}{i} \right |=\sum_{i=1}^{2n}\frac{1}{i} - \sum_{i=1}^{n}\frac{1}{i} = \sum_{i=n+1}^{2n}\frac{1}{i} = \frac{1}{n+1} +\frac{1}{n+2} + ... + \frac{1}{2n} < \frac{1}{n} +\frac{1}{n} + ... + \frac{1}{n} = n \left (\frac{1}{n} \right )=1 (=\varepsilon ) $$

But, remember that $\varepsilon >0$ is chose arbitrarily. So the harmonic series is not Cauchy.

(2) If $a_n$ is Cauchy, then $\lim_{n \to \infty} (x_{n+1}-x_n) = 0$. The converse is NOT true.

Counterexample for the converse:

Take $a_n=\sqrt{n}$.

$\lim_{n \to \infty} (\sqrt{n+1}-\sqrt{n}) = 0$, but $(\sqrt{n})_{n\in\mathbb{n}}$ is not Cauchy.


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