# Two-Sided Inequality - One Provenance

### Proof 1

Let $A=\displaystyle\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k}+\sqrt{2k+1}}\;$ and $B=\displaystyle\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k-1}+\sqrt{2k}}.\;$ Then

\displaystyle\begin{align} A &= \sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k}+\sqrt{2k+1}}= \sum_{k=1}^{2n(n+1)}\frac{\sqrt{2k+1}-\sqrt{2k}}{(2k+1)-(2k)}= \sum_{k=1}^{2n(n+1)}(\sqrt{2k+1}-\sqrt{2k})\\ &= \sum_{k=2}^{2n(n+1)+1}\sqrt{2k-1}-\sum_{k+1}^{2n(n+1)}\sqrt{2k},\\ B &= \sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k-1}+\sqrt{2k}}= \sum_{k=1}^{2n(n+1)}(\sqrt{2k}-\sqrt{2k-1})\\ &= \sum_{k=1}^{2n(n+1)}\sqrt{2k}-\sum_{k=1}^{2n(n+1)}\sqrt{2k-1}.\\ \end{align}

Now, note that, by telescoping,

\begin{align} A+B&=\sqrt{2(2n(n+1)+1)-1}-\sqrt{2\cdot 1-1}\\ &=\sqrt{4n^2+4n+1}-1\\ &=(2n+1)-1\\ &=2n. \end{align}

However, clearly $A-B\lt 0.\;$ Thus $A\lt n\lt B.$

### Proof 2

As before, define $A=\displaystyle\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k}+\sqrt{2k+1}}\;$ and $B=\displaystyle\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k-1}+\sqrt{2k}}.\;$ Then, by "telescoping",

\displaystyle\begin{align} A+B&=\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k}+\sqrt{2k+1}}+\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k-1}+\sqrt{2k}}\\ &=\sum_{k=1}^{2n(n+1)}(\sqrt{2k+1}-\sqrt{2k})+\sum_{k=1}^{2n(n+1)}(\sqrt{2k}-\sqrt{2k-1})\\ &=\sum_{k=1}^{2n(n+1)}(\sqrt{2k+1}-\sqrt{2k-1})\\ &=\sqrt{2\cdot 2n(n+1)+1}-1\\ &=(2n+1)-1\\ &=2n, \end{align}

so that $A+B=2n.\;$ Now, on one hand, $2A\lt A+B=2n,\;$ implying $A\lt n.\;$ On the other hand, $2B\gt A+B=2n,\;$ implying $B\gt n.$

### Acknowledgment

The problem has been kindly posted by Dorin Marghidanu and answered by Leonard Giugiuc at the CutTheKnotMath faceboook page. I find the problem elegant, the proof charming. A slightly different version of the proof has been submitted by Dorin Marghidanu.