Sum of fractions with square roots inequality What is the greatest integer $n$ such that
$n \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{2014}}$?
 A: 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.
A: We can view this as Riemann sum and compare with the integral (note that $x\mapsto\frac1{\sqrt x}$ is decreasing):
$$ 1+\int_1^{2014}\frac{\mathrm dx}{\sqrt x}>\sum_{k=1}^{2014}\frac1{\sqrt k}>\int_1^{2015}\frac{\mathrm dx}{\sqrt x}$$
i.e. 
$$ 88.76\approx2\sqrt{2014}-1>\sum_{k=1}^{2014}\frac1{\sqrt k}>2\sqrt{2015}-2\approx 87.8$$
Unfortunately, this is not enough to settle the question. But the lower bound can be improved to 
$$ \sum_{k=1}^{2014}\frac1{\sqrt k}>1+\frac1{\sqrt2}+\int_3^{2015}\frac{\mathrm dx}{\sqrt x}=1+\frac1{\sqrt2}+2(\sqrt{2015}-\sqrt 3)\approx 88.02$$
(I leave it to you if you compute $\sqrt{2015}$ etc. to sufficient precision or extract a few more of the initial summands; )
A: The integral criterium (see the last step of the proof of http://en.wikipedia.org/wiki/Integral_test_for_convergence) gives you some great upper and lower bounds:
$$\int_N^{M+1}f(x)\,dx\le\sum_{n=N}^Mf(n)\le f(N)+\int_N^M f(x)\,dx.$$
A: By using induction it's easy to see that
$$
2\sqrt{n+1}\lt2+\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\le2\sqrt{n}+1
$$
