Convergence of the sequence $a_n = \sqrt{n}$

Question

I know that in general, a Cauchy sequence isn't necessarily a convergent sequence. However, if the space the sequence is in is complete, all Cauchy sequences will converge.

Now the sequence given exists in the real numbers which are complete.

It's easy to see that $a_n \to \infty$ when $n \to \infty$, so this sequence cannot be convergent. Yet we can check that $(a_{n+1} - a_n) \to 0$ which would mean that it should be Cauchy since this should imply that $ \forall \epsilon > 0, \exists N$ s.t. If $m,n >N$ then $a_m - a_n < \epsilon$.

So on one hand it diverges, while on the other it's Cauchy in a complete metric space giving us convergence. What am I missing?

Answer 1

With $a_n=\sqrt{n}$, we have

$$\lim_{n\to\infty}(a_{4n}-a_n)=$$

$$\lim_{n\to\infty}\sqrt{n}=+\infty.$$

thus, $(a_n)$ is not a Cauchy sequence.