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My attempt at the proof goes as follows:

Suppose $(b_{n})$ is bounded. Then there exists $M>0 \in R$ such that for each $n \in N$ we are guaranteed that $|b_{n}|M$ by definition of floor. Then,

$|\frac{n^2}{n+1}|=|\frac{(\lfloor M\rfloor+1)^2}{\lfloor M\rfloor +2}|

But since $M>0$ by hypothesis, it follows that $|\frac{(\lfloor M\rfloor+1)^2}{\lfloor M\rfloor +2}|=\frac{(\lfloor M\rfloor+1)^2}{\lfloor M\rfloor +2}

$(\lfloor M\rfloor+1)^2

$\lfloor M\rfloor^{2}+2\lfloor M\rfloor+1

But $\lfloor M\rfloor + 1 > M$ and so the former is a contradiction. Hence, $(b_{n})$ is bounded. QED.

This is really the only argument that I could come up with, but I've never used the floor function prior to this. I believe I understand it, and I don't think I am committing any crime by using it. Let me know your thoughts, and please correct any mistakes I've made in this proof.

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    Wait, doesn't the remark that $b_n>n-1$ suffice to conclude? – 2017-02-14
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    I see where you are going, but show me. – 2017-02-14
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    What? $ $ $ $ $ $ – 2017-02-14
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    You said $b_{n}>n-1$ is a sufficient condition. I am not seeing the connection to what I am doing here. Please clarify. – 2017-02-14
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    Again: Can you show that $b_n>n-1$ for every $n$ implies directly that $b_n\to\infty$? If not, this is what you should be asking about... – 2017-02-14
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    @MrStormy83 Staying silent after my previous comment was perhaps not the wisest attitude. See how the posts accumulate below, with no way to know if any of them answers your query... – 2017-02-14
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    @Did sorry, I have a life outside of Stack Exchange. I've been working on other things and just came back to this today. – 2017-02-15
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    Right, so... your question arose from the trouble you had showing that $$n-1$$ is unbounded and an answer requiring to show that $$\frac{n}2$$ is unbounded, is what you were after? – 2017-02-15
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    My question arose from simply attempting to apply the definition of unbounded to prove that this sequence is unbounded. I picked the answer below because it helped me approach this proof in a completely different way than what I have above. In fact your advice helped as well so thank you. Sorry for being vague and not detailing my thought processes here, I realize that was misleading. In time I will hopefully learn to utilize this page better. – 2017-02-16

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An arguably simpler approach is to notice that for $n>1$, $2n > n+1$, hence $$\frac{1}{2}n = \frac{n^2}{2n} < \frac{n^2}{n+1}$$ You then need only show that the sequence $(n)_1^{\infty}$ is unbounded, which should be trivial.

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    They "then need only to show that the sequence" $(\tfrac12n)_1^\infty$ is unbounded, which might not be trivial to them since they seem to find difficult proving that $(n-1)_1^\infty$ is unbounded (see my comments on main). – 2017-02-14
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    To be fair, I too have no idea what part of the original post you were referring to by $b_n > n-1$. I'm not saying your approach is wrong, but I don't think the confusion is in the idea but the application. – 2017-02-14
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    @user153126 you are right, the confusion was in the application. – 2017-02-16
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Hint

$$\frac{n^2}{n+1}=\frac{(n+1)(n-1)+1}{n+1}=n-1+\frac{1}{n+1}$$

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$$b_n = \frac{n^2}{n+1} > \frac{n^2}{n+1} - \frac{1}{n+1} = \frac{(n+1)(n-1)}{n+1} = n-1,$$ so $\forall M > 0\ \forall N >0\ \exists n = \max\{\lfloor M\rfloor + 2, N + 1\} > N:\ b_n > M.$

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Edit: I was wrong the first time. But you should be fine showing that $\frac{n^2}{2n} \to \infty$ for $n \to \infty$. Since $\frac{n^2}{n+1} > \frac{n^2}{2n}$ the statement should follow.

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    $n^2\gt n+1$ isn't enough to show that $\lim_{n\to\infty}b_n\to\infty$. After all, $2n+1\gt n$ for all $n\gt0$, but $\lim\frac{2n+1}{n}\not\to\infty$. – 2017-02-14
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    You are right, let's edit this. – 2017-02-14
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    At the downvoter: Sorry for the first mistake. I saw and corrected it. Unfortunately not fast enough for you to downvote. Please give people enough time to correct her or his sloppyness. – 2017-02-14