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

There is a theorem in limits that says that if $\lim\limits_{n \to \infty} a_n$ exists then the sequence a_n is bounded.

And my professor gave me an example for this. Consider $a_n=\frac{1}{n}$ then $\lim\limits_{n \to \infty} a_n = 0$. The maximum bounds for this sequence is $1$ and the minimum bound is $0$.

Now instead if I take $a_n=\frac{1}{n-1}$ then $\lim\limits_{n \to \infty} a_n = 0$ which means the limit exists. Now though the sequence has a minimum bound of $0$ it has no maximum bound.

Now is there some sort of caveat in the theorem I'm missing?

Solution 1:

In general, a sequence is defined as a (real-valued) function on the naturals. $$a:\mathbb{N}\to\mathbb{R}$$ If you try to define a sequence by $a_n=\frac{1}{n-1}$, then you run into problems when you find that $a_1=\frac{1}{0}\notin\mathbb{R}$, meaning that this is not a valid sequence. Now, we could modify the definition above to allow sequences that start with $a_2$ to fix this, but that doesn't tell us what the real problem is here. The key thing to note is that each element of the sequence is a real number, which must be finite, which also tells us that any finite subset of the sequence must be bounded. The only way the sequence can be unbounded is if there is an infinite subsequence that goes to $\pm\infty$, which implies that the sequence does not converge.