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The standard proof of this theorem (Another similar question) is :

Proof:

Let $s_n$ be a convergent sequence, and let $\lim s_n = s$. Then for $\epsilon > 0$ we have:

$n > k \implies |s_n - s| < \epsilon$

This translates to : $\forall n > k , |s_n| < |s| + \epsilon$.

************** Area where I am confused *****************

Define $M= \max\{|s|+\epsilon, |s_1|, |s_2|, ..., |s_k|\}$. Then we have $|s_n| \leq M$ for all $n \in N$.

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What if there is a term that is infinite in the set between $\{|s_1|, |s_2|, ..., |s_k|)|\}$

Consider for example the sequence $(x_n)=\{1/(3-n)\}$.

We see that $$\lim_{n\to\infty} x_n = 0$$

Therefore this sequence converges to 0 but the sequence is not bounded as $|x_3| = \infty$

This sequence clearly contradicts the theorem. Is there a fault in my understanding of the definition a sequence or any other term?

Buddha
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    The sequence has to come from some set. If that set is the real or complex numbers, then $\infty$ is not in that set. Your element is not "infinite", it doesn't exist. – B. Goddard Sep 13 '22 at 11:26
  • @B.Goddard If I understand your explanation correctly the term 1/(3-3) is not a valid term as it is not defined ( I now see that its wrong to say that $x_3 = \infty$ rather its undefined). So the conclusion should be $x_n = {1/(3-n)}$ is a sequence in which one of the terms ($x_3$) is not valid and therefore is not a valid sequence? – Buddha Sep 13 '22 at 11:39
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    Correct, more or less. Wording is clunky, but you have the right idea. – Cameron L. Williams Sep 13 '22 at 12:24
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    There's a missing hypothesis in the statement of the theorem. It probably says, "Let $s_n$ be a sequence of real numbers." If one of your terms is not a real number, then the theorem can't apply. – B. Goddard Sep 13 '22 at 12:29

2 Answers2

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You are correct that the term $\frac{1}{3-3}$ is problematical. The best way to diagnose the problem, though, is not to say that this term is "infinite" but instead to simply say that it is undefined, because division by zero is undefined in the real numbers.

But in this case there is a workaround. Sequences are not always required to start at the index value $n=1$. It is fairly common to allow a sequence to start at a later value of the index. In this case, to get around the undefined term $\frac{1}{3-3}$, you can specify that the sequence is defined starting at the index value $n=4$. And once you've done that, you can proceed with applying the definition of convergence (but you'll have to fix the expression for $M$).

Lee Mosher
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Whenever you're talking about sequences it is important to clarify which set they're defined on. If you're taking a real analysis course it is mostly assumed that youre talking about sequences of real numbers. The sequence in your example is not a sequence of real numbers because it simply isn't defined for $n=3$.

Note that sometimes (for example in complex analysis) we adjoin the $\infty$ symbol to the complex numbers, so that we can explore poles of functions without worrying about the range. Sequences on $\mathbb C \cup \{\infty\}$ do make sense, however the term "bounded" loses it's meaning, because for that to exist you have to have a concept of distance, which by definition cant be infinite (there is more to this story, but that's besides the point). The key takeaway is: always make sure your sequence is well defined on the set in which it's supposed to lie.

Boxonix
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