Cauchy sequence and convergence - $ \frac {1}{n}$

Question

I have read that every convergent sequence is also a cauchy sequence and every cauchy sequence is convergent. I have found that the sequence given by $ \frac {1}{n}$ is Cauchy but $\sum_{i=1}^\infty \frac {1}{n}$ isn't obviously convergent because it has an infinite sum. I am confused. Is my misunderstanding caused by the fact we are just talking about the sequences, not the series ?

Answer 1

Let for $n>0$,

$$S_n=1+\frac 12+\frac 13+...\frac 1n. $$

we have

$$S_{2n}-S_n=\sum_{k=n+1}^{2n}\frac {1}{k}\ge n.\frac {1}{2n}=\frac 12$$

thus if $\epsilon=\frac 12$ then

$(\forall N\in \mathbb N)\;\; \exists p=N+1\;\; $ and $ \exists q=2 (N+1) : S_q-S_p\ge \epsilon $

hence $(S_n) $ is not Cauchy.

Answer 2

It is important to specify the space in which the points lie. The Cauchy criterion is equivalent to convergence to a limit if the underlying space is complete (the real numbers, for example, are complete). The sum $\sum_{n=1}^{\infty} \frac{1}{n}$ doesn't converge since the sequence $\{\sum_{n=1}^N \frac{1}{n}\}_{N=1}^{\infty}$ isn't Cauchy. The sequence $\{\frac{1}{n}\}_{n=1}^{\infty}$ is Cauchy and indeed converges to $0$.