Christmas Streak 05/88: Too Many Terms!

Note that $\sqrt{40\pm 2\sqrt{n}}=\sqrt{20+\sqrt{400-n}}\pm\sqrt{20-\sqrt{400-n}}.$

$\begin{aligned} &\frac{\displaystyle \sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\displaystyle \sum_{k=1}^{399} \sqrt{20-\sqrt{k}}} \\\\ & = \frac{\displaystyle \sum_{k=1}^{399} \sqrt{40+2\sqrt{k}}}{\displaystyle \sum_{k=1}^{399} \sqrt{40-2\sqrt{k}}} \\\\ & = \frac{\displaystyle \sum_{k=1}^{399} \left(\sqrt{20+\sqrt{400-n}}+\sqrt{20-\sqrt{400-n}}\right)}{\displaystyle \sum_{k=1}^{399} \left(\sqrt{20+\sqrt{400-n}}-\sqrt{20-\sqrt{400-n}}\right)} \\\\ & = \frac{\displaystyle \sum_{k=1}^{399} \left(\sqrt{20+\sqrt{n}}+\sqrt{20-\sqrt{n}}\right)}{\displaystyle \sum_{k=1}^{399} \left(\sqrt{20+\sqrt{n}}-\sqrt{20-\sqrt{n}}\right)} && \small \color{#3D99F6} \text{Note that }n\text{ and }400-n\text{ are symmetric when }n=1,~2,~3,~\cdots,~399. \\\\ \end{aligned}$

Let $\displaystyle \sum_{k=1}^{399} \sqrt{20+\sqrt{k}}=p$ and $\displaystyle \sum_{k=1}^{399} \sqrt{20-\sqrt{k}}=q.$

Then, from above,

$\begin{aligned} & \frac{p}{q} =\frac{p+q}{p-q} \\\\ & p^2+2pq-q^2 =0 \\\\ & \left(\frac{p}{q}\right)^2+2\left(\frac{p}{q}\right)-1= 0 \\\\ & \frac{p}{q} = 1+\sqrt{2}~(\because~p>0,~q>0) \end{aligned}$

Therefore, the original expression is equal to $1+\sqrt{2}=2.414213562\cdots\approx\boxed{2.414}.$

I just did again,

$\displaystyle \dfrac{\int_0^{399} \sqrt{20 + \sqrt{x}}}{\int_0^{399} \sqrt{20 - \sqrt{x}}}$ and I obtained 2.407851555085965002023512067819406583363069582927281754428 . XD

Anyways. Nice solution H.M. :D