I have a statement where I have to construct a sequence in which the statement makes sense and make an example of the sequence and then say whether it is true or false:
The statement is: If a sequence is eventually increasing, then it is bounded below.
I have to determine whether the statement is true or false, and say why...
How would I go about determining whether it is true or false, and constructing a sequence to show my workings on how i got it to be true/false?
What you need first is some intuition. You need to formulate your best guess as to whether it's true or false. Try thinking about what eventually increasing sequences look like. What can you do with them? Can you make them have no lower bound? Could you make them, say, tend to $-\infty$?
If nothing springs to mind try, playing around with drawing graphs of some sequences (just puts some dots on a page), making sure that they eventually start increasing. See what's possible. Then you might be able to see whether it's true or false.
Once you've done this, hopefully you can see that it is indeed true, and that you have some intuitive idea why it must be true. It's then a matter of converting this intuitive idea into a formal proof.
After playing with these sequences enough, hopefully you can see that the minimum value (yes, not just bounded below, but you should be seeing a minimum value) for the sequence must happen at or before the point it starts increasing. That makes sense, right? Because everything beyond that point will be at least as large as the sequence value where the sequence starts increasing!
Let's start bringing in definitions. We say a sequence is eventually increasing if, after some point $N$ (for all $n \ge N$), each term of our sequence is greater than the previous one, that is, $$n \ge N \implies x_{n+1} \ge x_n.$$ So, if each term is larger than the previous one, this tells us that $$x_N \le x_{N+1} \le x_{N+2} \le x_{N+3} \le \dots$$ That is, $x_N$ is a lower bound on the tail of the sequence, starting at $N$. Does this mean $x_N$ is a lower bound on the whole sequence? Hopefully your examples show this is not necessarily the case. We have to consider all the previous terms of the sequence too. Can they be a problem? Not at all, since there's only finitely many of them! They must have a minimum (and a maximum too, though that's not relevant).
So, hopefully it should be reasonable to you that our lower bound should be
$$l = \min \lbrace x_1, x_2, \ldots, x_N \rbrace.$$
How do we prove it? Well, let's appeal to the definition of lower bound: $l$ is a lower bound for $(x_n)$ if $l \le x_n$ for all $n$. We already know that, if $n \ge N$, then $x_n \ge x_N$, and using the definition of the minimum, $x_N \ge l$, so $x_n \ge l$ as required. If $n < N$, then $x_n$ appears in the set $\lbrace x_1, x_2, \ldots, x_N \rbrace$, so using the definition of minimum again, $l \le x_n$. Either way, $l \le x_n$ and $l$ is a lower bound for the sequence.