4

Since the definition of Cauchy sequence is: Understanding the definition of Cauchy sequence, I noticed we need an absolute value for $a_m-a_n$ in the definition so the statement would be false. But I can't find such a couterexample. Maybe it is true?

Community
  • 1
Maggie Mak
  • 339
  • 1
    No, it is not true. –  Jan 30 '16 at 20:24
  • 1
    But instead if you have something like $x_{n+1} - x_n = a_n$ where $\sum {a_n}$ exists, then the sequence would be cauchy – Mambo Jan 30 '16 at 20:27
  • 2
    A very stupid (yet good as rule of thumb): if it were true, why would one bother with a more complicated definition for Cauchy sequences? – Clement C. Jan 30 '16 at 20:51
  • See http://math.stackexchange.com/questions/55735/pseudo-cauchy-sequences-are-they-also-cauchy and http://math.stackexchange.com/questions/768404/boundedness-and-cauchy-sequence-is-a-bounded-sequence-such-that-lima-n1-a – Martin Sleziak Mar 18 '16 at 06:39
  • See also https://math.stackexchange.com/questions/2891922 – Watson Jul 06 '20 at 14:26
  • https://math.stackexchange.com/questions/410787 – Watson Aug 20 '20 at 09:51

5 Answers5

8

No, for the same reason that some series don't converge, even though their terms go to zero. In fact, consider the sequence

$$a_n = 1 + \frac12 + \dotsb + \frac{1}{n}.$$

The difference between its terms goes to zero, but the sequence is not Cauchy (it goes to infinity).

Eric Auld
  • 28,127
7

A very simple example: $(u_n=\sqrt n)$ is such that $$u_{n+1}-u_n=-\frac 1{\sqrt n+\sqrt{n+1}}\to 0$$ but $(u_n)$ tends to $+\infty$, and thus can't be a Cauchy sequence.

Bernard
  • 175,478
5

Hint Consider the sequence $$x_n := \sum_{i = 1}^n \frac{1}{i} .$$

Travis Willse
  • 99,363
5

The harmonic series is a counterexample: $x_{n+1}-x_{n}=\frac{1}{n} \to 0$, but $x_{n}$ does not converge. The 'moral' is that Cauchy sequences give you control over the tail in a kind of stronger way than just knowing successive terms are close - they tell you the whole tail is eventually close (which, in a complete space, tells you it converges).

preferred_anon
  • 17,121
4

Not quite. Think of the following sequence: $$ 0,1,\frac{1}{2},0,\frac{1}{3},\frac{2}{3},1,\frac{3}{4},\frac{2}{4},\frac{1}{4},0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1,... $$ In words, we move back and forth from $0$ to $1$. Each time we "turn around," i.e. reach a $0$ or $1$, our step size decreases. What can you say about $\lim_{n\to\infty}x_{n+1}-x_n$? What can you say about whether or not the sequence converges/is Cauchy?

TomGrubb
  • 12,909
  • This is a nice example, since (as opposed to the other examples), this sequence is bounded. – Watson Jul 06 '20 at 14:22