# Sum involving Triangular Numbers

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 $$\frac{99}{\sqrt[4]{8}}$$

• I like your answer (+1) Commented Aug 12, 2023 at 5:59