If a sequence converges then the sequence is bounded? I am taking an online real analysis class. In the class on sequences and series, the professor said the above statement is true since when a series converges to a point it is bounded in its neighborhood.
But 1/(n-2) is clearly a series that converges to 0 but is not bounded in the neighborhood of 2. (considering n to be a rational number). Since many pointed out that the series is not defined at 2. I am defining it to be 0 for n=2. So it seems like I am missing something here. Or is the above statement false?
Here is the link to the video https://www.youtube.com/watch?v=VEx3Ys6JAJo&list=PL0E754696F72137EC&index=15
 A: You seem to be confusing the definition of a sequence. A sequence is a countable list of real numbers (possibly finite or infinite). Thats it. It has a $1$ term, a $2$ term, a $3$ term, and so on.
When you say: what about the sequence $\frac{1}{n-2}$ for $n\in\mathbb{N}$, at $n=2$? The answer is that this is not a sequence. In fact, it is a sequence for $n\geq 3$, but you cannot call an undefined value as part of a sequence.
But you say, what about the sequence $\frac{1}{n-2}$ for all $n\in\mathbb{R}^+$ except for $n=2$? You are correct, this function is unbounded around $n=2$. However, a sequence takes as inputs natural numbers, not real numbers. Thus, what you have described is again not a sequence.
I think a main point you are misunderstanding is that generally, $n$ is taken to be a natural number. That is, $n\in\mathbb{N}$. It is sloppy notation to define a sequence as $a_n=\frac{1}{n-2}$ without also saying what happens at $n=2$. However, mathematicians will generally just ignore this undefined term (or let it be $0$).

But you say, what if you let $n$ run over all rational numbers except for $2$? Well, then what we are really doing is defining
$$a_n=\frac{1}{b_n-2}$$
where $b_n$ is any enumeration of the rationals except for $2$. But then this sequence does not converge. Let $M>0$ be given. Then there is some $k$ such that
$$b_k=\frac{4\lceil M\rceil+1}{2\lceil M\rceil}$$
Then
$$a_k=\frac{1}{b_k-2}=\frac{1}{\frac{4\lceil M\rceil+1}{2\lceil M\rceil}-2}=2\lceil M\rceil>M$$
Thus, the sequence does not converge. That is, the sequence definately converges (ignoring the issue at $n=2$) if $n$ goes over the naturals. However, if $n$ runs over some other countable set then the sequence may not converge.
A: Suppose $s_{n}$ converges to $s$. Let $\epsilon > 0$ be given. Then there exists $N$ such that $|s_{n} - s| < \epsilon$ for all $n \geq N$. This means $s_{n} \in (s - \epsilon, s + \epsilon)$ for all $n \geq N$.
A: Here it is the standard proof of such result.
Proposition
Let $x_{n}\in\mathbb{R}$ be a sequence which converges to $x_{0}\in\mathbb{R}$, then it is bounded.
Proof
Let $\varepsilon = 1$. Then, according to the definition of limit, there exists $n_{1}\in\mathbb{N}$ such that
\begin{align*}
n\geq n_{1} \Rightarrow |x_{n} - x_{0}| < 1 \Rightarrow |x_{n}| - |x_{0}| < 1 \Rightarrow |x_{n}| < |x_{0}| + 1
\end{align*}
Thus, if we take $M = \max\{|x_{1}|,|x_{2}|,\ldots,|x_{n_{1}-1}|,|x_{0}| + 1\}$, we can conclude that $|x_{n}| \leq M$.
That is to say $x_{n}$ is bounded, and we are done.
Hopefully this helps!
