# 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.

• So I guess you don't like the idea that the sequence $1,2,5,10,1,1,1,1,1,\cdots$ might converge to $1$?
– user239203
Aug 21, 2019 at 17:10
• Also, your definition is completely oblivious of the fact that a sequence should converge to some number. It's like saying that "a convergent sequence is a sequence such that every five terms there is a positive one". Good for you, but what's the distinguishing feature between any two such sequences?
– user239203
Aug 21, 2019 at 17:24
• No constant sequence satisfies you condition. Aug 21, 2019 at 17:31

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.

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.

• The sequence $0,1,0,0,0,0,\dots$ doesn't converge?
– zhw.
Aug 21, 2019 at 17:20
• I wrote the opposite of what I meant. What do you think now? Aug 21, 2019 at 17:26

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 $$a_n\to 0.$$