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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}

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.

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    Take $(a_n)$ to be a decreasing sequence bounded below by some $a>0$. Following your definition, $(a_n)$ converges to every number in $(-\infty,a]$. – 2017-01-07
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    @OpenBall please see the edited answer. – 2017-01-07
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    Here's two fundamental problems for any definition along these lines: first of all, there are strictly decreasing sequences which don't tend to zero (e.g. $1.1, 1.01, 1.001, 1.0001, ...$); if the $D_n$s form such a sequence, then that will satisfy your definition, but won't correspond to a convergent sequence. Second, note that **the constant sequence $(a, a, a, ...)$ fails your condition $D_n>D_{n+1}$!** So by your definition, the constant sequence doesn't converge. – 2017-01-07

2 Answers 2

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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}

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}

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.

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    please see the edited answer. – 2017-01-07
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    @ShreyAryan I have edited my answer, but if you want to change your definition again, I suggest you to check your new definition with diverse examples of sequences and to ask a new question on this site afterwards. Don't edit your question, because then the answers won't match. :) – 2017-01-07
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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.

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    Please read the edited answer. – 2017-01-07
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    The example still is a counter example. It is always useful to construct examples with the even and odd subsequences, so define $a_{2m}=1+\frac1{2m}$ and $a_{2m+1}=1+\frac1m$ to even destroy the monotonicity of the example to show that none of your properties are essential for convergence. – 2017-01-07