Doubt regarding definition of converging sequence. I am currently studying about converging sequence in my Real Analysis class.
The definition of a converging sequence is

A sequence of real numbers converges to a real number a if, for
  every positive number $\epsilon$, there exists an $N \in \mathbb N$ such that for all $n ≥ N$, $|a_n - a| < \varepsilon$.

However instead of using $\varepsilon$ why can't we just define it as follows.

A sequence of real numbers converges to a real number if, $\forall\ n\ \in \mathbb N$ and $n > 2$, $|a_n-a_{n-1}| < |a_{n-1}-a_{n-2}|$.

Any help would be appreciated.
 A: I think the other folks who've commented and answered have done a good job explaining why you can't use the explicit definition you've given, but I also think your basic intuition is in the right place. The definition you've given is the standard one, but there is also the idea of a Cauchy sequence in real analysis, which is what I think you are trying to aim at. 
Specifically, if you check out Rudin's Principles of Mathematical Analysis (3rd Ed.), he proves that every sequence in $R^k$ converges if and only if it is Cauchy. See theorem 3.11 in that work.
A: No, we can't. For instance, if $a_n=\sum_{k=1}^n\frac1k$, then the condition that you stated holds, but $(a_n)_{n\in\mathbb N}$ diverges.
On the other hand, the sequence $0,1,0,0,0,0,\ldots$ converges, but your condition doesn't holds for it.
A: An equivalent def'n of $A=\lim_{n\to \infty}a_n$ is that whenever $r>0 ,$ the set $\{n\in \Bbb N: a_n\not \in (A-r,A+r)\}$ is finite. Verbally we can say this as:  Each  neighborhood of $A$ (no matter how small) contains $a_n$ for all but finitely many $n\in \Bbb N$. 
This does not require $a_n$ to progress in any "orderly fashion" towards $A$ as $n$ increases. For example if $a_n=2^{-n}$ when $n$ is odd and $a_n=10^{-n}$ when $n$ is even then $a_n\to 0.$
