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

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 ?

• Indeed $\{1/n\}$ is a Cauchy sequence and converges to $0$. But the sequence of partial series $\{\sum_{i=1}^N 1/n\}$ is NOT a Cauchy sequence. Commented Jun 21, 2017 at 0:41
• Oh, I understand it now. I was confused how the sequence can be made up by series, but it's made by the partial sums $s_1, s_2, ...., s_n$, right?
– Leif
Commented Jun 21, 2017 at 0:43

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.

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 $0$.

• So I just take the limit of $\frac {1}{n}$ and see that it goes to an zero, so it's convergent and Cauchy. But if I have the sum of $\frac {1}{n}$, it doesn't converge since the sum must be finite?
– Leif
Commented Jun 21, 2017 at 0:34