Let $T_n = \frac{n(n+1)}{2}$, the $n$th triangular number. What is the sum $$\sum_{n=1}^{9999}\sqrt{\sqrt{T_n+\frac{1}{8}}-\sqrt{T_n}}\hspace{0.4cm}?$$
I have tried simplifying the expression inside the outer square root by substituting $A=T_N+\frac{1}{16}$ to get $\sqrt{A+\frac{1}{16}} - \sqrt{A-\frac{1}{16}}$ and hopefully remove some square roots, but that didn't yield anything meaningful.
Substituting $T_n=\frac{n(n+1)}{2}$, the summand is equal to \begin{align*} &\sqrt{\sqrt{\frac{n(n+1)}{2}+\frac{1}{8}}-\sqrt{\frac{n(n+1)}{2}}}\\ =&\frac{1}{\sqrt[4]{8}}\sqrt{\sqrt{4n(n+1)+1}-2\sqrt{n(n+1)}}\\ =&\frac{1}{\sqrt[4]{8}}\sqrt{\sqrt{4n^2+4n+1}-2\sqrt{n(n+1)}}\\ =&\frac{1}{\sqrt[4]{8}}\sqrt{2n+1-2\sqrt{n(n+1)}}\\ =&\frac{1}{\sqrt[4]{8}}\sqrt{n+(n+1)-2\sqrt{n(n+1)}}\\ =&\frac{1}{\sqrt[4]{8}}\sqrt{\left(\sqrt{n+1}-\sqrt{n}\right)^2}\\ =&\frac{1}{\sqrt[4]{8}}\left(\sqrt{n+1}-\sqrt{n}\right)\\ \end{align*} And now, we can see the sum telescopes to $\frac{99}{\sqrt[4]{8}}$
