# 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

$$x_{n}:=\Biggr(\dfrac{n^2}{n+1}\Bigg)^{n=\infty}$$

• I assume that you mean the limit as $n\to\infty$, not as $x\to\infty$. – Brian M. Scott Aug 29 '13 at 4:21
• Ok I'm wrong I will fix it – Hugus Aug 29 '13 at 4:34

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.)

• This is a very nice suggestion – Hugus Aug 29 '13 at 4:38

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 $n\to\infty$.

HINT: $\quad n^2=(n+1)(n-1)+1$.

$$\frac{n^2}{n+1}=\frac{\frac{n^2}{n}}{\frac{n}{n}+\frac{1}{n}}=\frac{n}{1+\frac{1}{n}}$$ As $n\to\infty$, we diverge.

• How I can forget this !! :) – Hugus Aug 29 '13 at 4:39
• You shouldn't.... ;) – Eleven-Eleven Aug 29 '13 at 4:49

Just notice that,

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

Given $N$, $\quad\exists\ N'\ \ni\ n > N' \quad\Longrightarrow\quad n^{2}/\left(n + 1\right) > N$

$$\mbox{where}\qquad N' = \left\lfloor {N + \sqrt{\vphantom{\LARGE a}\,N\left(N + 4\right)\,} \over 2} \right\rfloor\,, \qquad N' \in {\mathbb N}$$