If the limit of a sequence exists then the sequence is bounded

I had earlier asked a question to which I did not get a satisfactory answer. But then if I think of it, I feel like I could have thrown some more light on the problem I'm facing.

If $\lim\limits_{n \to \infty} a_n$ exists is it true that $a_n$ is bounded?

My professor stated that if a limit of a sequence exists, then the sequence exists.

To prove it, he said if $\lim\limits_{n \to \infty} a_n = L$, where $L \in \mathbb{R}$

$\forall \varepsilon \gt 0 \ \exists N$, such that all the elements in the sequence greater than N would certainly exist in the open interval $(L-\varepsilon, L+\varepsilon)$.

So what this implies is the elements in the sequence outside the above interval contribute to the bounds of the sequence.

which implies the maximum bound on $a_n$ is $max(L+\varepsilon,a_1,a_2,.....,a_N)$ and minimum bound is $min(L-\varepsilon,a_1,a_2,.....,a_N)$.

So if I consider the sequence $a_n = \frac{1}{n}$ which has a finite limit $\lim\limits_{n \to \infty} \frac{1}{n} = 0$, and if I take an $\varepsilon \gt \frac{1}{4} \gt 0$ , so the maximum bound would be $max(\frac{1}{4},\frac{1}{2},1)=1$ and minimum would be $0$.

the same cannot be said for $a_n = \frac{1}{n-1}$ which also has a finite limit $\lim\limits_{n \to \infty} \frac{1}{n-1} = 0$ but is definitely not bounded.

So is this because $a_n=\frac{1}{n-1}$ or $a_n=\frac{1}{n-M}$ where $M \in \mathbb R$ for that matter is not a valid sequence?

Solution 1:

A sequence $(a_n)_{n\in \mathbb{N}}$ in $X$ is a function

$a:\mathbb{N}\to X$ where we denote $a(n) $ by $a_n$.

In a metric space $(X, d) $ every convergent sequence is bounded.

In your question, $a_n=\frac{1}{n-1}$ is defined for all $n\ge 2$.

And the sequence , $(a_n)_{n\ge 2}=(1,1/2,1/3,...) $

And it is clearly bounded above by 1 and bounded below by 0.