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I have to show that the sequence $x_n = \displaystyle\frac{n}{\sqrt{n+1}}$ is properly divergent from the definiton of proper divergence. This is the following definition I must use.

" For any $\alpha\in\mathbb{R}$, there exits $K(\alpha)\in\mathbb{N}$ such that if $n\geq K(\alpha)$ then $x_n> \alpha$. This implies $\lim\limits_{n\to\infty}x_n = \infty$

Some help would really be appreicated!

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For any $\alpha$, for any $n>(|\alpha|+1)^{2}$, then \begin{align*} \dfrac{n}{\sqrt{n+1}}&=\dfrac{n+1}{\sqrt{n+1}}-\dfrac{1}{\sqrt{n+1}}\\ &=\sqrt{n+1}-\dfrac{1}{\sqrt{n+1}}\\ &>\sqrt{n+1}-1\\ &>\sqrt{(|\alpha|+1)^{2}}-1\\ &=|\alpha|+1-1\\ &=|\alpha|\\ &\geq\alpha. \end{align*}

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