Prove the divergence of the sequence $ \lim_{x\to\infty}\frac{n^2}{n+1} $ I am looking for nice ways of proving the divergence of this sequence
$$x_{n}:=\Biggr(\dfrac{n^2}{n+1}\Bigg)^{n=\infty}$$  
 A: If you use polynomial long division, you can rewrite $\frac{n^2}{n+1}$ as $n-1+\frac{1}{n+1}$. That should make it more clear that the terms keep growing as $n\to\infty$.
A: In the spirit of Brian M. Scott's answer, observe that:
$$n^2 - 1 < n^2 < n^2 + 2n + 1$$
So we have:
$$\frac{n^2 - 1}{n+1} < \frac{n^2}{n+1} < \frac{n^2 + 2n + 1}{n+1}$$
That is,
$$n-1 < x_n < n+1$$
Since $n-1$ and $n+1$ both (clearly) diverge to infinity as $n \rightarrow \infty$, the same can be said of $x_n$. 
(I suppose you only need the left inequality; but at least this lets you see what the behavior of $x_n$ looks like.)
A: HINT: $\quad n^2=(n+1)(n-1)+1$.
A: $$\frac{n^2}{n+1}=\frac{\frac{n^2}{n}}{\frac{n}{n}+\frac{1}{n}}=\frac{n}{1+\frac{1}{n}}$$
As $n\to\infty$, we diverge.
A: Just notice that,

$$ n+1 \sim n \quad \rm as \quad n\to \infty .$$ 

A: Given $N$, $\quad\exists\ N'\ \ni\ n > N'
\quad\Longrightarrow\quad n^{2}/\left(n + 1\right) > N$
$$
\mbox{where}\qquad
N'
=
\left\lfloor
{N + \sqrt{\vphantom{\LARGE a}\,N\left(N + 4\right)\,}
 \over
 2}
\right\rfloor\,,
\qquad
N' \in {\mathbb N}
$$ 
