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To show that the given sequence is decreasing, show that ${a}_{n}\ge {a}_{n\+1}$, a result that can be established by the following (seemingly magical) calculations. The secret to the origins of these calculations is revealed at the end.
$n\left(n\+1\right)$

$\ge 1$

${n}^{2}\+n$

$\ge 1$

$\left({n}^{3}\+{n}^{2}\right)\+{n}^{2}\+n$

$\ge 1plus;\left({n}^{3}plus;{n}^{2}\right)$

$n\left({\left(n\+1\right)}^{2}\+1\right)$

$\ge \left({n}^{2}plus;1\right)\left(nplus;1\right)$

$\frac{n}{{n}^{2}\+1}$

$\ge \frac{nplus;1}{{\left(nplus;1\right)}^{2}plus;1}$

${a}_{n}$

$\ge {a}_{nplus;1}$



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The starting inequality and the terms added to both sides in the third step seem to come from nowhere. However, it would be bad mathematics to start with the final inequality, which is not known to be true, and manipulate that. However, that is just what is done to divine the "magic" steps that need to be invoked. At least one popular calculus text suggests "cross multiplying" the inequality to be proved and working down to the inequality that starts the calculation above. In fact, this device can also be found in other math texts, but there, a phrase something like the cryptic "the steps are reversible, so the result has been proven" might be appended.
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The function $f\left(x\right)\=\frac{x}{{x}^{2}\+1}$ could more easily be shown to be decreasing by showing that $f\prime \left(x\right)\=\frac{{x}^{2}1}{{\left({x}^{2}\+1\right)}^{2}}$ is clearly negative for $\leftx\right\>1$. Consequently, since ${a}_{n}\=f\left(n\right)$, that the given sequence is decreasing is more easily established by this device as by the manipulation of inequalities.
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