How should Theorem 3.22 in Baby Rudin be modified so as to yield Theorem 3.23 as a special case? I'm reading Walter Rudin's PRINCIPLES OF MATHEMATICAL ANALYSIS, 3rd edition, and am at Theorem 3.22. 
Theorem 3.22: Let $\{a_n\}$ be a sequence of complex numbers. Then $\sum a_n$ 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 $m=n$, this becomes $$\left|a_n \right| \leq \epsilon$$ if $n \geq N$. 
In other words, 
Theorem 3.23: If $\sum a_n$ converges, then $\lim_{n\to\infty} a_n = 0$. 
The condition $\lim_{n\to\infty} a_n = 0$ is not, however, sufficient to ensure convergence of $\sum a_n$. For instance the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges although $1/n \to 0$ as $n \to \infty$. 
The above information is what I've copied almost verbatim from Rudin. 
Now I have a couple of questions:
Why does the condition in Theorem 3.22 has used $\geq \epsilon$ instead of $< \epsilon$ as is the case with Theorem 3.11 (or more precisely Definition 3.9) from which Theorem 3.22 stems? 
And, in going from Theorem 3.22 to Theorem 3.23 by taking the special case when $m=n$, why is it that the double implication is lost? Is this not a pitfall? If not, why? And if it is a pitfall, then should the statement of Theorem 3.22 not be modified to avoid this pitfall? 

*

*In this kind of proof "something $\leq \epsilon$" it does not matter "$\leq$" or "$<$", since it carries something like "for any sufficiently small $\epsilon$".

*The double implication is lost since $m$ can be much larger than $n$, and if "$a_n$ does not decreases fast enough", you cannot control $\sum_{k=n}^m a_k$ using $a_n$.
