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:

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

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?

• wolframalpha.com/input/… – David Peterson Aug 15 '15 at 7:11
• @DavidP: So is this a rational number??? – barak manos Aug 15 '15 at 7:13
• highly unlikely wolframalpha.com/input/… – David Peterson Aug 15 '15 at 7:15
• @DavidP That's actually $\sqrt 2 + 1$ – Deepak Aug 15 '15 at 7:23
• It looks as if $\dfrac{\displaystyle\sum_{n=1}^{k^2-1} \sqrt{k + \sqrt{n}}}{\displaystyle\sum_{n=1}^{k^2-1} \sqrt{k - \sqrt{n}}} = \sqrt{2}+1$ for integer $k\gt 1$ – Henry Aug 15 '15 at 7:30

Hint: Show that for all relevant $n$ $$\frac{\sqrt{10+\sqrt{100-n}}+\sqrt{10+\sqrt{n}}}{\sqrt{10-\sqrt{100-n}}+ \sqrt{10-\sqrt{n}}}=\sqrt2 +1.$$ IOW pair up terms in the numerator and the denominator starting from both ends.

[Edit:]

Claim. Assume that $a,b,c$, all positive, are the lengths of the sides of a right angled triangle - a Pythagorean triple if you like - $c$ is the hypotenuse. Then $$\frac{\sqrt{c+a}+\sqrt{c+b}}{\sqrt{c-a}+\sqrt{c-b}}=1+\sqrt2.$$

Proof. The left hand side of the claim is clearly immune to scaling. We can adjust the scale so that $c-b=2$. Then a calculation (familiar to enthusiasts of Pythagorean triples) shows that for some positive real number $m$ we have $$c=m^2+1,\quad b=m^2-1,\quad a=2m.$$ (IOW instead of the usual integer parametrization in terms of $(m,n)$ we set $n=1$, and let $m$ be arbitrary.) We then see that the numerator is $m+1+\sqrt2 m=m(1+\sqrt2)+1$, and the denominator is $m-1+\sqrt2$. Because $(\sqrt2+1)^{-1}=\sqrt2-1$ the claim follows. Q.E.D.

Leaving it to the reader to derive the identity of my hint as a corollary of the claim.

• It goes too hairy to show that ratio. Is there any other way around? – frukoprof Aug 15 '15 at 7:46
• Not sure about the best way. One way is to use a trigonometric substitution $n=100\sin^2x$ and work with half-angle formulas. That's how I verified it. The ratio I wrote simplifies to $$\frac{\cos t+\cos(\pi/4-t)}{\sin t+\sin(\pi/4-t)}=1+\sqrt2$$ for $t=x/2$ Admittedly that feels like hiding the key point. :-( The appearance of the formulas somehow suggested trig substitution for me. May be a geometric justification is better?! – Jyrki Lahtonen Aug 15 '15 at 7:58
• Actually this ratio known as Silver Ratio, as you know. Therefore as you say, a geometric representation can be helpful. I'll give it a try. If you get something helpful please comment here. Thank you :) – frukoprof Aug 15 '15 at 8:13
• I do not know how to go from your hint the the solution of the problem. What is amazing is that the derivative of your expression with respect to $n$ is strictly $0$ (tedious calculation). So, the rhs can be computed for $n=0$ to get the result !. – Claude Leibovici Aug 15 '15 at 8:15
• Summary of how I feel. My "big" contribution was to observe that we should pair up the terms like here. I am leaving the search for the best way of proving constancy of this ratio for others. It's the last weekend of my vacation, and I want to go to the forest and forage a bucketload of blueberries (and fight mosquitoes and flies). <Homer Simpson voice> Mmmm...Blueberry pie. </Homer Simpson voice> – Jyrki Lahtonen Aug 15 '15 at 8:23

For the Calculation of $$\displaystyle \frac{\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}} =$$

Let $$\displaystyle A_{n} = \sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}$$ and $$\displaystyle B_{n} = \sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}$$ , where $n>1$

Now $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right)^2 = 2n-2\sqrt{n^2-k}$$

So $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$$

So $$\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$$

So So $$\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{k}}$$

So $$A_{n}-B_{n} = B_{n}\sqrt{2}$$

So $$A_{n} = B_{n}\left(1+\sqrt{2}\right)$$

So $$\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$$

Now Put $\displaystyle n^2-1 = 99\Rightarrow n= 10\;,$ So we get $$\displaystyle \frac{\sum_{k=1}^{99}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{99}\sqrt{n-\sqrt{k}}} =\frac{A_{10}}{B_{10}} = 1+\sqrt{2}.$$

• Very nice and clever! (+1) – Markus Scheuer Aug 15 '15 at 18:13
• This is nice. Note that it is not necessary to have a perfect square minus one as the maximum. The same arguments show that e.g.$$\frac{\sqrt{\sqrt{2016}+\sqrt{1}}+\sqrt{\sqrt{2016}+\sqrt{2}}+\cdots+\sqrt{\sqrt{2016}+\sqrt{2015}}} {\sqrt{\sqrt{2016}-\sqrt{1}}+\sqrt{\sqrt{2016}-\sqrt{2}}+\cdots+\sqrt{\sqrt{2016}-\sqrt{2015}}}=1+\sqrt2.$$ – Jyrki Lahtonen Aug 15 '15 at 18:23
• Yes Jyrki Lahtonen You are Right, – juantheron Aug 15 '15 at 18:27