Find value of $\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$

Question

Find value of

$$S=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}$$

I started with $$S+\frac{1}{S}=\frac{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}+\frac{\sum_{k=1}^{399} \sqrt{20-\sqrt{k}}}{\sum_{k=1}^{399} \sqrt{20+\sqrt{k}}}$$

$\implies$

$$S+\frac{1}{S}=\frac{\sum_{i=1}^{399}(20+\sqrt{i})+2S_1+\sum_{i=1}^{399}(20-\sqrt{i})+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$ $\implies$

$$S+\frac{1}{S}=\frac{15960+2S_1+2S_2}{\left(\sum_{k=1}^{399}\sqrt{20+\sqrt{k}}\right) \times \left(\sum_{k=1}^{399}\sqrt{20-\sqrt{k}}\right)}$$

where $$S_1=\sum_{i \ne j=1}^{399}\left(\sqrt{20+\sqrt{i}}\right)\left(\sqrt{20+\sqrt{j}}\right)$$ and like wise

$$S_2=\sum_{i \ne j=1}^{399}\left(\sqrt{20-\sqrt{i}}\right)\left(\sqrt{20-\sqrt{j}}\right)$$

Any way to proceed from here?

Answer 1

Let

$$S_1=\sum_{k=1}^{m^2-1} \sqrt{m+\sqrt{k}}$$

and

$$S_2=\sum_{k=1}^{m^2-1} \sqrt{m-\sqrt{k}}.$$

(your sum is $S_1/S_2$ at $m=20$). Note that

$$\left(\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}\right)^2=2m+2\sqrt{m^2-k},$$

so

$$\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}=\sqrt{2}\sqrt{m+\sqrt{m^2-k}}.$$

Thus

$$S_1+S_2=\sum_{k=1}^{m^2-1} \left(\sqrt{m+\sqrt{k}}+\sqrt{m-\sqrt{k}}\right) = \sqrt{2}\sum_{k=1}^{m^2-1} \sqrt{m+\sqrt{m^2-k}}.$$

However, under the substitution $k\to m^2-k$, the sum on the RHS is simply $S_1$. Thus,

$$S_1+S_2=S_1\sqrt{2}$$

$$S_2=\left(\sqrt{2}-1\right)S_1$$

$$\frac{S_1}{S_2}=\boxed{1+\sqrt{2}}.$$

As Robert Israel points out, this result is independent of $m$.

Answer 2

Amazingly, it appears that $$ \dfrac{\sum_{k=1}^{m^2-1} \sqrt{m + \sqrt{k}}}{\sum_{k=1}^{m^2-1} \sqrt{m - \sqrt{k}}} = 1 + \sqrt{2} $$ for all integers $m \ge 2$.