Criteria for convergence in Rudin

Question

Theorem 3.22: $\sum a_n $ converges if and only if for every $\epsilon >0$ there is an integer $N$ such that $$ \left|\sum_{k=n}^m a_k \right | \leq \epsilon $$ if $ m \geq n \geq N $

In particular, by taking $m=n$ becomes $$ \left|a_n \right | \leq \epsilon ~~~~~ (n\geq N)$$

However, the condition $ \left|a_n \right | \leq \epsilon $ is satisfied by $a_n = \frac{1}{n}$. Thus by the backward implication of the theorem, $\sum \frac{1}{n} $ should converge. However, as we know it, it does not.

Which part of the theorem have I misunderstood?

Edit:

Would this be the correct interpretation of the theorem? $$\sum a_n~converges \iff (\forall\epsilon>0)(\exists N \in \mathbb{N})\left( m \geq n \geq N \implies \left|\sum_{k=n}^m a_k \right | \leq \epsilon \right) $$

Answer 1

Yes, it is true that, for every $\varepsilon>0$, we have $\frac1n<\varepsilon$ if $n$ is large enough. But you can't deduce from thath fact that$$\sum_{k=n}^m\frac1k<\varepsilon$$if $m\geqslant n$ and $m$ and $n$ are large enough.

Answer 2

In other words, let $$S_n=\sum_{k=0}^na_k$$

the criteria is

$$\sum a_n\text{ converges } \iff (S_n) \text{ is Cauchy} \iff$$

$$\forall \epsilon>0 \;\; \exists N\in \Bbb N :$$

$$m>n\ge N \implies |S_m-S_n|<\epsilon$$

Answer 3

If $m\ge n\ge N$ then $\left|\sum_{k=n}^ma_k\right|\le\epsilon$

means for all $m$,$n$ such that $m\ge n\ge N$. Taking $m=n$ is not equivalent to convergence.