Convergence implies boundness of a sequence but an example confuses me!

Question

given the following sequence

$$ p_{n} = \frac{1}{n-1} $$

I am tempted to say that for n=1 the sequence assumes an infinite (and thus unbounded) value but still the sequence converges to zero as n increases.
I am of course wrong because this contradicts theorem 3.2c of Baby Rudin which states that "if a sequence is convergent in a metric space then it's bounded". What am I missing? Maybe 1/0 is not a mathematical defined value? I would be greatly thankful for an answer.

Erma

For $n=1$, $p_n$ is not defined. or if you say that one of the sequence terms is infinite, that term is certainly not a number, and so the sequence is not a sequence of numbers — Hagen von Eitzen, Sep 18 '20 at 20:38
What is the set of values the index is taking? Is $n=1$ actually in the set of such values? — Eric Towers, Sep 18 '20 at 20:38
@EricTowers Yes I am considering n=1,2,3,... Sorry for being not precise about that. — erma, Sep 19 '20 at 06:59

user · Answer 1 · 2020-09-18T20:49:36.020

-1

Single values for which $p_n$ is not defined are not relevant when we take the limit as $n\to \infty$ what matters is that $p_n$ is eventually a well defined expression.

For example

$$\lim_{n\to \infty} \frac1{\sqrt{n^2-1\,000\,n-1\,000\,000}}$$

is not well defined for a finite number of values but the limit is equal to $0$ since eventually the expression inside the square root becomes positive and tends to $\infty$.

edited Sep 18 '20 at 20:49

answered Sep 18 '20 at 20:37

user

154,566

This is interesting, thanks. – erma Sep 19 '20 at 07:00
@erma You are welcome! Bye – user Sep 19 '20 at 07:03

Convergence implies boundness of a sequence but an example confuses me!

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