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In investigating of convergence of a sequence we use $n\longrightarrow \infty $ . why we can only use $\infty$ and we can not use the other numbers for convergence in a sequence as convergence of the other functions?

thank you.

joker
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    How would you understand $\lim\limits_{n\to 5} a_n$ for a sequence? Keep in mind that a sequence is a function $\mathbb N\to M$ where $M$ is a non-empty set, other than with a function e.g. $f:\mathbb R\to\mathbb R$ there simply is no such thing as a limit $n\to 5$. – Hirshy May 15 '16 at 17:38
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    Also, $\infty$ here should be treated only as a symbol, not as a number. I.e., we might have introduced a notation $\operatorname{LIM}na_n$ instead of $\lim{n\to\infty}a_n$ because no reference to a mystical object $\infty$ is needed to define the limit of a sequence (or even the limit of a map with any countably infinite domain and values in a topological space) – Hagen von Eitzen May 15 '16 at 17:47
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    @Hirshy There is -- it'll just be $a_5$. – Clement C. May 15 '16 at 18:04
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    @ClementC. Any real number you want would be a limit – la flaca May 15 '16 at 18:52
  • @Eliana True -- my bad. – Clement C. May 15 '16 at 19:22
  • I prefer \to which gives $\to$ instead of \rightarrow which gives $\rightarrow$. Less typing. – DanielWainfleet May 16 '16 at 16:06

2 Answers2

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Recall the definition:

We say that the limit of a function $f:M_1 \to M_2$ from one metric space to another is $L$ when $x \to a$ $(x,a$ elements of $M_1)$whenever for any $\epsilon>0$ (a positive real number), there exists some $\delta>0$ (another positive real number) such that for any $x \in M_1- \lbrace a \rbrace$ it is the case that $d_1(x,a)< \delta \implies d_2(f(x),L) < \epsilon$, where $d_1$ and $d_2$ are the distance functions of $M_1$ and $M_2$ respectively.

Now, if you want to apply this definition to a function of the form $f: \mathbb{N} \to M$ (a sequence of elements of a metric space, the complex if you want) and you consider $\mathbb{N}$ as a metric space with the restriction of the usual metric of the reals to $\mathbb{N}$ (which means that the distance between $m,n \in \mathbb{N}$ is $|m-n|$), the problem is that the function doesn't "get as close as you want" to any particular value when you stay within some neighbourhood of some $a \in \mathbb{N}$ because its image is a discrete set.

Take for instance the function $f:\mathbb{N} \to \mathbb{R}$ defined as $f(n)=\frac{1}{1+n}$ and suppose you want to know which is the limit as $n \to 2$, then you will get that any real number $L$ counts as a limit of the function because the distance between two diferent natural numbers is always greater or equal than $1$, so any $0 <\delta \leq 1$ will make the statement: $$\forall \epsilon >0 \exists \delta >0 \forall n \in \mathbb{N}- \lbrace 2 \rbrace [|n-a|< \delta \implies |f(n) - L|< \epsilon]$$ be true. Hence the notion of a limit of a sequence when $n$ approaches some natural $a$ becomes useless.

la flaca
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This is a good question and rarely is this fact emphasized in textbooks of calculus/real-analysis. So +1 goes for OP.

The idea behind limits is the fact it is used to analyze the behavior of a function $f$ near a point $c$ (say) (which may or may not lie in domain of the function). Moreover the fundamental part of analyzing the behavior of functions in such manner is that there should be a point in the domain of $f$ which is as near to $c$ as we please. Thus by induction there should be infinitely many points in domain of $f$ which are as near to $c$ as we want. It is therefore essential that the domain of $f$ must be an infinite set in order for limit operations to work.

In case of sequences the domain of the function is $\mathbb{N}$ and it is a discrete domain where the distance between any two points of the domain is at least $1$. So if you take a point $c$ which is finite, you are not going to get infinitely many points of domain close to $c$. The only option left is to consider the behavior of the sequence / function for large values of its argument and then we have no end to supply of points of the domain of $f$ which are near to $\infty$. The concept of points near $\infty$ roughly gets translated into points which are as large as we please.

Paramanand Singh
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