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Any converging sequence is bounded.

This is a theorem in our book. It is proved by picking the first (finite) number of elements and show that the others are smaller than some constant. But, I can't agree with this theorem. For example, $a_n=\frac{1}{n-1}$ is not bounded (by $n=1$) (or is it)?

Regards

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Did you mean to write $n > 1$? If yes then $a_n$ is bounded by 1. –  Matt N. Oct 28 '11 at 16:35
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The sequence $a_n$ is not defined at $n=1$. –  Sasha Oct 28 '11 at 16:36
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The theorem assumes (correctly so) that $a_1$ is a real number first. In your example, $a_1$ is $\frac{1}{0}$ which is not a meaningful expression at all. –  Srivatsan Oct 28 '11 at 16:36
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@Kevin, there's a tacit convention that one doesn't care too much about whether the defining expression fails to make sense for the first few elements in the series, as long as the properties one is interested in are ones that cannot be affected by modifying finitely many elements of a sequence. Being bounded and being convergent are both properties of this kind, so the division by zero at $n=1$ would usually just be ignored without comment. We can just take the sequence to be defined for $n\ge 2$, which then has all the properties we're interested in. –  Henning Makholm Oct 28 '11 at 17:39
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@Kevin: Or to put Henning's comment another way: A sequence is supposed to be a function with domain $\mathbb{N}$. The formula you give makes no sense for $n=1$, so the formula you give does not define a function with domain $\mathbb{N}$; you have to either give an explicit definition for what your sequence will be at $n=1$, or make some modification so that you get a sequence (or change the meaning of "sequence", which one can also do). In any case, you never consider "$1/0$" when dealing with that sequence. –  Arturo Magidin Oct 28 '11 at 19:22

2 Answers 2

up vote 1 down vote accepted

The definition of a convergent sequence is a sequence $a_n$ which has a limit $a$. We also know that:

$$\lim_{n\to\infty} a_n=a\iff\forall\varepsilon>0\exists N\in\mathbb N\forall n>N\colon|a_n-a|<\varepsilon$$

Note that as the comments indicate, we can always replace the first $k$ elements, for any given $k$, since we only care about what happens from some point.

Taking a limit is what happens after the fat lady sings, in some sense. It is an asymptotic behavior of a sequence, it occurs when the sequences eventually moves towards a certain point.

From the above definition it should be clear why every convergent sequence is bounded. For only finitely many elements are bigger than $a+\dfrac{1}{2}$, where $a$ is the limit of the sequence.

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When you use a induction, you need follow the sentence below:

$(\forall P)[[P(0) \land ( \forall k \in \mathbb{N}) (P(k) \Rightarrow P(k+1))] \Rightarrow ( \forall n \in \mathbb{N} ) [ P(n) ]]$

In other words, you need prove to $P(0)$ and prove if works to $P(k)$ works to $P(k+1)$ too.

In your example, $P(k) \Rightarrow P(k+1)$ holds, but you $P(0)$ no, because, $a_1=\frac{1}{1-1}$ don't exists.

But for a convergence sequence (note $a_1$,$a_2$,$a_3$,..., isn't a convergent sequence, because, $a_1$ isn't not greater than $a_2$, $a_1$ is indefinite) that $P(0)$ and $P(k) \Rightarrow P(k+1)$ holds the theorem are ok.

Note too, $a_2,a_3,a_4,...$ is a convergence sequence and bounded.

PS.: A good propostion P to this case is ask, is P(k) limitated?

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