Is this definition of convergence of a sequence equivalent to the actual definition? Book-ish definition:

A sequence $(a_n)$ converges to a real number $a$ if, for every positive number $\epsilon$, there exists an $N\in\mathbb{N}$ such that whenever $n\geq N$ it follows that $|a_n-a|< \epsilon.$

My Definition: (Edit)
A sequence $(a_n)$ converges to a real number $a$ if there exists an $N\in \mathbb{N}$ such that for all $n\geq N$ if $D_n=|a_n-a|$ then $D_{n+1}<D_n$ and the set $S=\{D_n\}$ does not contain a lower bound.  
Note: I am beginning to learn real analysis by trying to first guess the definitions and theorems myself before accepting the ones laid out by the author and so if I am wrong please correct me (but do so gently). Also please comment on whether this way of learning analysis is the right way to do so. 
 A: As you edited your question, I have to edit my answer.
Initial answer before the OP changed their definition
Your initial definition was:

A sequence $(a_n)$ converges to a real number $a$ if there exists an $N\in \mathbb{N}$ such that for all $n\geq N$ if $D_n=|a_n-a|$ then $D_{n+1}<D_n.$ 

EDIT2: First of all, I assumed that you meant $D_{n}\geq D_{n+1}$ or it is obviously a problem for any constant sequence.
Take the following sequence for example:
$$a_{n}:=
\begin{cases}
\frac{1}{n} &\text{if } n \text{ even}\\
0 &\text{if } n \text{ odd }
\end{cases}$$
We see that for any $n$ odd, we have $D_{n}=\vert a_{n}-0\vert=0<\vert a_{n+1}-0\vert = \frac{1}{n+1}=D_{n+1}$.
Yet, the sequence $(a_{n})_{n\in\Bbb N}$ converges to $0$.
Nevertheless, trying to figure out useful definitions is a good exercise to learn by oneself because it challenges the mind and makes you active in the learning process: you are not just understanding what is written in the book, in a somehow passive way.
EDIT after the OP changed their definition
Your second definition:

A sequence $(a_n)$ converges to a real number $a$ if there exists an $N\in \mathbb{N}$ such that for all $n\geq N$ if $D_n=|a_n-a|$ then $D_{n+1}<D_n$ and the set $S=\{D_n\}$ does not contain a lower bound.

EDIT2: First of all, I assumed that you meant $D_{n}\geq D_{n+1}$ or it is obviously a problem for any constant sequence.
It still doesn't work but here in an obvious way: the set $\{D_{n}\}$ is always bounded below by $0$ by its very definition! If you mean that the lower bound cannot exist or belong to the set $S=\{D_{n}\vert n\geq N\}$, then it fails if the sequence is constant, which is problematic.
Please, don't edit your question again: ask a new one.
A: Try some examples like $a_n=1+\frac1n$ and $a=0$ to find the failings of your statement.
Of course, $a_n\to a$ is equivalent to $D_n=|a_n-a|\to 0$, but that convergence is not always monotonous.
