More fun in 2014-2015

$\begin{aligned} S & = \sum_{n=1}^{2014} \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}} \\ & = \sum_{n=1}^{2014} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{((n+1)\sqrt{n} + n\sqrt{n+1})((n+1)\sqrt{n} - n\sqrt{n+1}} \\ & = \sum_{n=1}^{2014} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)^2 - n^2(n+1)} \\ & = \sum_{n=1}^{2014} \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)} \\ & = \sum_{n=1}^{2014} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right) \\ & = \sum_{n=1}^{2014} \frac{1}{\sqrt{n}} - \sum_{n=2}^{2015} \frac{1}{\sqrt{n}} \\ & = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2015}} \\ & = \frac{\sqrt{2015} -1} {\sqrt{2015}} \end{aligned}$

$\approx 1$

Another way. We clearly see that every fractions have this form $\dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$ $=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}$ $=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}$ Replace this form into A and we have $A=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dots + \dfrac{1}{\sqrt{2014}}-\dfrac{1}{\sqrt{2015}}$ $A=1-\dfrac{1}{\sqrt{2015}}\approx{0.977}$ Which is approximately 1

$A=\frac{1}{\sqrt{1}\times \sqrt{2}(\sqrt{2}+\sqrt{1})}+\frac{1}{\sqrt{2}\times \sqrt{3}(\sqrt{3}+\sqrt{2})}+...+\frac{1}{\sqrt{2014}\times \sqrt{2015}(\sqrt{2014}+\sqrt{2015})}$

$=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{1}\times\sqrt{2} }+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}\times \sqrt{3}}+...+\frac{\sqrt{2015}-\sqrt{2014}}{\sqrt{2014}\times \sqrt{2015}}$ $=(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}})+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}})+...+(\frac{1}{\sqrt{2014}}-\frac{1}{\sqrt{2015}})$ $=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2015}}=1-\frac{1}{\sqrt{2015}}$ $\approx 1$