Compute ​

This sum is obviously telescoping :)

\(\begin{array} {rll} \phantom{=}&\displaystyle\sum^{9999}_{n=1}\dfrac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\color{red}{\sqrt[4]{n+1}-\sqrt[4]{n}}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)\color{red}{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{n+1 - n}\\ =&\displaystyle \sum^{9999}_{n=1}\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)\\ =&\displaystyle (\color{blue}{\sqrt[4]{2}}\color{black} - \sqrt[4]1) + (\color{green}{\sqrt[4]{3}}\color{black} - \color{blue}\sqrt[4]2\color{black}) + (\sqrt[4]4 - \color{green}\sqrt[4]3\color{black}) + \cdots+(\sqrt[4]{10000}-\sqrt[4]{9999})\\ =&\displaystyle \sqrt[4]{10000} - \sqrt[4]1\\ =&\displaystyle 10 - 1\\ =& 9 \end{array}\)

sum_(n=1)^9999 1/((sqrt(n) + sqrt(n + 1)) (n^(1/4) + (n + 1)^(1/4)))≈9.000000000000000000000000000000000000000