Show sequence $a_n = \frac{n^2}{n+1}$ is divergent 
Show sequence $a_n = \frac{n^2}{n+1}$ is divergent

My thinking is, we shall show $a_n$ is unbounded, and hence cannot be convergent.
Suppose $M \in \mathbb{R}$ is an upper bound for $a_n$.
Choose $N \in \mathbb{R}$ so that $N > 2M$ (Do I need to justify why I can do this? If so how?)
Notice, for all $n > N$:
$$\frac{n^2}{n+1} > \frac{n^2}{n+n} = \frac{n^2}{2n} = \frac{n}{2} > \frac{N}{2} > M$$
Hence we cannot find an upper bound for $a_n$, and so not convergent.
 A: Two minor comments.
One, very minor:
You can justify the existence of $N$ by recalling the Archimedean property of integers.

Two, slightly less minor:
You don't need to phrase your proof as a proof from contradiction. In fact, it is usually preferred to not do so if possible, because proofs by contradiction are usually slightly less elegant and slightly harder to follow. Your proof can be rewritten ever so slightly like so:

Let $M\in\mathbb R$. Then, there exists $N \in \mathbb{R}$ so that $N > 2M$
Notice, for all $n > N$:
$$\frac{n^2}{n+1} > \frac{n^2}{n+n} = \frac{n^2}{2n} = \frac{n}{2} >
\frac{N}{2} > M$$ and therefore, $M$ is not the upper bound of the sequence. Since $M$ was arbitrary, we conclude the sequence does not have an upper bound.

This is true even more in general. Many many proofs by contradiction, especially by less experienced students, are not really proofs by contradiction, and most of the time, they can be rewritten into a more direct form.
As a rule of thumb, you can spot them because they look something like this:

*

*We are trying to prove that the statement $\exists x\in A: P(x)$ is not true.

*Let $x\in A$.

*Assume $P(x)$ is true.

*Something something.

*Therefore, $\neg P(x)$.

*This is a contradiction with the assumption $P(x)$.

*Therefore, $\neg P(x)$.

*Therefore, $\forall x\in A: \neg P(x)$, qed.

You might notice there is a bit of duplication in the proof above. We are concluding $\neg P(x)$ twice. And sometimes we are. However, be careful, it's very important what happens in the "something something" stage of the proof. There are two options:

*

*If the assumption $P(x)$ is used inside the somethingsomething part of the proof, then the proof is inherently a proof by contradiction. An example of this would be the standard proof of infinitude of prime numbers. In that proof, we specifically use the assumption that there is only a finite number of primes, and then construct a new number that is coprime to all of them.*

*However, the cases I am describing are such that the assumption $P(x)$ is not used in the somethingsomething part of the proof. In that case, the same proof can be rewritten simply as:


*

*We are trying to prove that the statement $\exists x\in A: P(x)$ is not true.

*Let $x\in A$.

*Something something.

*Therefore, $\neg P(x)$.

*Therefore, $\forall x\in A:\neg P(x)$, qed.

* Now, technically, even that proof can be rewritten into a more constructive form, but in that case, the rewrite would be very extensive and would not improve readability.
