Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:

$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$

(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?

Let $f(n)=\frac{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}+\sqrt{k}}}{\sum_{k=1}^{n} \sqrt{\sqrt{n+1}-\sqrt{k}}}$, and the original problem is to calculate $f(99)$.
Now we claim that $f(n)=\sqrt{2}+1$ for $\forall n>0$.
$f(n)=\frac{\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}} \cdot \frac{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}+\sqrt{k})(\sqrt{2}-1)}}{\sum_{k=1}^{n} \sqrt{(\sqrt{n+1}-\sqrt{k})(\sqrt{2}+1)}}$
$=(\sqrt{2}+1) \cdot \frac{\sum_{k=1}^{n} \sqrt{A(k)+B(k)}}{\sum_{k=1}^{n} \sqrt{A(k)-B(k)}}$, where $A(k)=\sqrt{2(n+1)}-\sqrt{k}$, $B(k)=\sqrt{2k}-\sqrt{n+1}$.
Notice that:
$\left(\sqrt{A(k)+B(k)}-\sqrt{A(k)-B(k)}\right)^2=2A(k)-2\sqrt{A(k)^2-B(k)^2}\\=2\left(\sqrt{2(n+1)}-\sqrt{k}-\sqrt{n+1-k}\right)$
That means 
$\left(\sqrt{A(n+1-k)+B(n+1-k)}-\sqrt{A(n+1-k)-B(n+1-k)}\right)^2\\=2\left(\sqrt{2(n+1)}-\sqrt{n+1-k}-\sqrt{k}\right)$
So that 
$\sqrt{A(k)+B(k)}-\sqrt{A(k)-B(k)}\\=\sqrt{A(n+1-k)-B(n+1-k)}-\sqrt{A(n+1-k)+B(n+1-k)}$
So that 
$\sum_{k=1}^{n} \sqrt{A(k)+B(k)}=\sum_{k=1}^{n} \sqrt{A(k)-B(k)}$
So that $f(n)=\boxed{\sqrt{2}+1}$
