If $\{x_n\}$ satisfies that $x_{n+1} - x_n$ goes to $0$, is $\{x_n\}$ a Cauchy sequence? [duplicate]

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?

Solution 1:

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).

Solution 2:

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.

Solution 3:

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

Solution 4:

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).

Solution 5:

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?