# Thread: Cauchy sequences

1. ## Cauchy sequences

Let {x sub n} be a sequence and suppose that the sequence {x sub (n+1) - x sub n} converges to 0. Give an example to show that the sequence {x sub n} may not converge. Hence, the condition that abs(x sub n-x sub m) is less than sigma for all m, n greater than or equal to N is crucial in the definition of a Cauchy sequence.

2. Originally Posted by friday616
Let {x sub n} be a sequence and suppose that the sequence {x sub (n+1) - x sub n} converges to 0. Give an example to show that the sequence {x sub n} may not converge. Hence, the condition that abs(x sub n-x sub m) is less than sigma for all m, n greater than or equal to N is crucial in the definition of a Cauchy sequence.
How about the sequence in $\mathbb{Q}$: 1, 1.4, 1.41, 1.414, 1.4142...

The idea of this sequence is that if it were in $\mathbb{R}$, it would converge to $\sqrt{2}$; however, since $\sqrt{2}\notin\mathbb{Q}$, this sequence doesn't converge to anything, though the points do get arbitrarily close together. (You know the deal: $\forall~\epsilon>0$, $\exists~N$ such that $n,m>N$ implies that $|x_n-x_m|<\epsilon$)

3. Originally Posted by friday616
Let {x sub n} be a sequence and suppose that the sequence {x sub (n+1) - x sub n} converges to 0. Give an example to show that the sequence {x sub n} may not converge. Hence, the condition that abs(x sub n-x sub m) is less than sigma for all m, n greater than or equal to N is crucial in the definition of a Cauchy sequence.
Let $x_n:= \sum_{j=1} ^{n} \ \frac{1}{j}$ then $x_{n+1} - x_n = \frac{1}{n+1} \rightarrow 0$ but $x_n$ doesn't converge

4. That's a good one.

5. How about $x_n = log(n)$.

We have $x_{n+1} - x_n = log(n+1) - log(n) = log(\frac{n+1}{n}) \rightarrow 0$.