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I am looking for nice ways of proving the divergence of this sequence

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

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

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  • $\begingroup$ This is a very nice suggestion $\endgroup$ – Hugus Aug 29 '13 at 4:38
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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 $n\to\infty$.

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

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

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    $\begingroup$ How I can forget this !! :) $\endgroup$ – Hugus Aug 29 '13 at 4:39
  • $\begingroup$ You shouldn't.... ;) $\endgroup$ – Eleven-Eleven Aug 29 '13 at 4:49
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Just notice that,

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

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

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

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