<s>[INST] Sum of fractions with square roots inequality What is the greatest integer $n$ such that
$n \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{2014}}$?
 [/INST] For positive integer $k$, we have $$2\sqrt{k} < \sqrt{k+1} + \sqrt{k} < 2\sqrt{k+1}$$
Taking reciprocals and rationalising the middle term, we get
$$\frac1{2\sqrt{k+1}}< \sqrt{k+1}-\sqrt k < \frac1{2\sqrt k}$$
The middle term telescopes and hence sums easily.  Thus we can have tight bounds on both sides:
$$\sum_{k=1}^{2014}\frac1{\sqrt{k}} = 1+\sum_{k=1}^{2013}\frac1{\sqrt{k+1}} < 1+2\sum_{k=1}^{2013} (\sqrt{k+1}-\sqrt{k})= 2\sqrt{2014}-1 \approx 88.75$$ and
$$\sum_{k=1}^{2014} \frac1{\sqrt k} > 1+\frac1{\sqrt2}+2\sum_{k=3}^{2014}(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{2015} - 2\sqrt3+1+\frac1{\sqrt2} \approx 88.02 $$
so you have your integer.
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