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Show that the 3-tail of the sequence defined by $x_n := \frac{n}{n^2+16}$ is monotone decreasing. Hint: Suppose $n \geq m \geq 4$ and consider the numerator of the expression $x_n-x_m$.

I am a bit confused of how to do this problem. My thinking is that to show that anything is monotone decreasing, we need to show that $x_n \geq x_{n+1} \forall n \in \mathbb{N}$. Is it correct to just show $x_n := \frac{n}{n^2+16}$ is monotone decreasing, because that implies the tail is also monotone decreasing? How does the hint help me?

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Let $x\gt 4$, $x$ real.

$f(x):=\dfrac{x}{x^2+16}$.

$f'(x)= \dfrac{(x^2+16) - x(2x)}{(x^2+16)^2}$;

$f'(x) =\dfrac {-x^2 +16}{(x^2+16)^2}.$

Note:

$f'(x) \lt 0$ for $x>4$, which implies

$f$ is decreasing for $x>4.$

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Hint: Show that $a_n:= {n^2+16\over n} = n+{16\over n}$ is increasing.


$$ a_{n+1}-a_n = 1-{16\over n(n+1)} >0$$ if $n>3$.

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  • $\begingroup$ I thought of approaching the problem in this way but the hint given seems to suggest to do something else. $\endgroup$ – Madhav Nakar Feb 21 '18 at 11:59
  • $\begingroup$ Well perhaps you can notice that in the numerator is quadratic function and this one is increasing much faster than linear function, so everything must decrese. $\endgroup$ – Aqua Feb 21 '18 at 12:03
  • $\begingroup$ Christian . Nice +. $\endgroup$ – Peter Szilas Feb 21 '18 at 13:10

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