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The exercise says:

Show that the sequence $x_n = \frac{n-3}{\sqrt{n^2+1}}$ is monotone beginning from some number $n \in \mathbb{N}$.

By definition a sequence is nondecreasing if $\forall n \in \mathbb{N}$ the following inequality holds

$$x_{n+1} \geq x_n$$

I'm able to show that $x_{n+1} > x_n$ (strict inequality) for all $n \geq 1$, but not after raising both sides of the last inequality to $2$ in order to get rid of the square roots and getting a quite messy expression.

My question is, how can this be done in a neater/more elegant way?

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$f(x) = \frac{x-3}{\sqrt{x^2+1}}$ is an increasing function for $x > 0$, since $$f'(x) = \frac{3x+1}{(x^2+1)^{3/2}}$$ Hence $f(n) < f(n+1)$ for all $n \geq 1$. The graph of $f(x)$ is shown below: enter image description here

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  • $\begingroup$ $\lim_{n\rightarrow \infty} f(n) = 1$ $\endgroup$ – user348749 Sep 16 '16 at 6:22
  • $\begingroup$ I also noticed that (it's like you're using geogebra like me (: ). But I'm trying just to prove the monotonicity without analyzing its derivative. $\endgroup$ – Jazz Sep 16 '16 at 6:44

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