Prove that if a sequence diverges to infinity then it is bounded below. This question was asked in our math test. What I did was assume that a divergent sequence is not bounded below. Then we would not know if the series diverges or converges as it would seem to diverge from both sides. Please help.
 A: Suppose that the sequence $x_1,x_2,\ldots$ diverges to infinity. By definition, this means that for any $M\in\mathbb R$, there exists $N\ge1$ such that $x_n>M$ for each $n>N$. It follows that
$$
x_n\ge\min\{x_1,\ldots,x_{N},M\}
$$
for each $n\ge1$. This means that the sequence is bounded from below.
A: Since the series diverges to $\infty$, there is an $n_0$, such that $a_n\ge 0$ for all $n\ge n_0$. 
With m:=min{$\ a_1,\cdots ,a_{n_0-1},0\ $} , we have $a_n \ge m$ for all $n$
A: You have the right idea, but unless you are able to formalize it, you don't have an answer yet.
Now, you want to prove it by contradiction, and that is possible. But you need something to hold on to. So, first off, name your sequence. Call it $(x_n)_{n=1}^\infty$.
Now, look at the definition of the two properties you have, that is "converging to $\infty$" and "bounded from below".

A sequence $(x_n)_{n=1}^\infty$ converges to infinity if, for every $M\in\mathbb R$, there exists some $N\in\mathbb N$ such that $x_n>M$ for $n>N$.

and

A sequence is $(x_n)_{n=1}^\infty$ is bounded from below if there exists some $m\in\mathbb R$ such that $x_n>m$ for all $n\in\mathbb N$.

OK, but you have, by assumption, a sequence that is not bounded from below. So, first, you have to write down what it means (formally) that a sequence is not bounded from above. (Hint: It should start with something like "for all $m\in\mathbb R$, there exists some $n\in\mathbb N$).
Once you have that figured out, you can look at the two properties your sequence is supposed to have and how they contradict each other.
