Convergence iff distance equals zero The following excerpt has been taken from Rudin's Principles of Mathematical Analysis:

If $d(p_n,p) < \epsilon$ for all $\epsilon$ than $d(p_n,p) = 0$. Why isn't $d(p_n,p) = 0$ written in the textbook rather than $d(p_n,p) < \epsilon$?
 A: If $d(p_{n},p) < \varepsilon$ for all $\varepsilon > 0$ and all $n$, then $p_{n} = p$ for all $n$, so $p_{n} \to p$ trivially. But then only constant sequences can converge, and therefore convergence theory gets real boring!
What the definition of convergence requires is not that stringent; it only requires that, for every $\varepsilon > 0$, there is some $N \geq 1$ such that $d(p_{n},p) < \varepsilon$ for all $n \geq N$. In other words, it requires that, for every $\varepsilon > 0$, we can obtain $d(p_{n},p) < \varepsilon$ from some term on. In this case, several sequences are by definition convergent then. In fact, the definition of convergence does not look at the "finite history" of a sequence. It forgives all the "inappropriate" behavior of a sequence finitely many times and, as long as the sequence behaves "right" eventually, the realm of convergent sequences counts it in. 
A: According to your definition, $\lim\limits_{n\to\infty}\frac{1}{n}$ would not exist.
Pay attention to what it's saying, it's saying that for every $\epsilon$ there is some $N$ such that $n>N$ implies $d(p_n,p)<\epsilon$. That does NOT mean that $d(p_n,p)<\epsilon$ for every $n$ or even for ANY $n$. 
