# 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/… Aug 15, 2015 at 7:11
• @DavidP: So is this a rational number??? Aug 15, 2015 at 7:13
• highly unlikely wolframalpha.com/input/… Aug 15, 2015 at 7:15
• @DavidP That's actually $\sqrt 2 + 1$ Aug 15, 2015 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 $k\gt 1$ Aug 15, 2015 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? Aug 15, 2015 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?! Aug 15, 2015 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 :) Aug 15, 2015 at 8:13
• 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> Aug 15, 2015 at 8:23
• @JyrkiLahtonen +1 just for priorities. :) Aug 15, 2015 at 8:31

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) Aug 15, 2015 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.$$\frac{\sqrt{\sqrt{2016}+\sqrt{1}}+\sqrt{\sqrt{2016}+\sqrt{2}}+\cdots+\sqrt{\sqrt{2016}+\sqrt{2015}}} {\sqrt{\sqrt{2016}-\sqrt{1}}+\sqrt{\sqrt{2016}-\sqrt{2}}+\cdots+\sqrt{\sqrt{2016}-\sqrt{2015}}}=1+\sqrt2.$$ Aug 15, 2015 at 18:23
• Yes Jyrki Lahtonen You are Right, Aug 15, 2015 at 18:27

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 $$f(n)=\boxed{\sqrt{2}+1}$$

• No problems! Thanks for trying to correct me Oct 22, 2019 at 17:03
• This answer deserves more upvotes. It doesn't seem like people realize you've proved a more general theorem than the others, because $n$ needn't be a square. Oct 22, 2019 at 17:23
• Thanks for the consideration! Oct 22, 2019 at 17:33
• If I did it with a colleague, it's not just mine Oct 22, 2019 at 17:34

The hint $$\frac{ \sqrt{1+\sin(x)}+\sqrt{1+\cos(x)} }{ \sqrt{1-\sin(x)}+\sqrt{1-\cos(x)} }=1+\sqrt2 \quad \text{for } x\in [0,\pi/2]$$ is spot on.

We have \begin{align} \sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{99}} &= \sqrt{10}\left(\sqrt{1+\sqrt{0.01}}+\sqrt{1+\sqrt{0.99}}\right) \\&= \sqrt{10}\left(\sqrt{1+\sin(t)}+\sqrt{1+\cos(t)}\right) \\&= \sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sin(t)}+\sqrt{1-\cos(t)}\right) \\&= \sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sqrt{0.01}}+\sqrt{1-\sqrt{0.99}}\right) \\&= (1+\sqrt 2)\left(\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{99}}\right) \end{align} for some $$t$$. Therefore, we can rearrange the numerator to be $$1+\sqrt 2$$ times the denominator. (You need to handle the middle terms separately but it works out the same.)

• Interesting (+1). I assume you can generalize this to $$\frac{\sum_{n=1}^{m^2-1}\sqrt{m+\sqrt{n}}}{\sum_{n=1}^{m^2-1}\sqrt{m-\sqrt{n}}}=1+\sqrt{2}?$$ Oct 22, 2019 at 16:07
• @clathratus, yes, it seems pretty straightforward.
– lhf
Oct 22, 2019 at 16:08
• In the middle (whiche happens in general for even $m$), you need one case where you do not pair two summands, but of course $\frac{\sqrt{10+\sqrt{50}}}{\sqrt{10-\sqrt{50}}}=1+\sqrt 2$ as well. Oct 22, 2019 at 16:20
• @HagenvonEitzen, yes, of course. Thanks for point it out.
– lhf
Oct 22, 2019 at 16:20

Let the numerator and the denominator

$$N= \sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\ldots+\sqrt{10+\sqrt{99}}$$ $$D =\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\ldots+\sqrt{10-\sqrt{99}}$$

Apply the denesting formulas

$$\sqrt{a\pm\sqrt c} = \frac{1}{\sqrt2} \left( \sqrt{a+\sqrt{a^2-c}} \pm \sqrt{a-\sqrt{a^2-c}} \right)$$

to get

$$\sqrt{10\pm\sqrt n} = \frac{1}{\sqrt2} \left( \sqrt{10+\sqrt{100-n}} \pm \sqrt{10-\sqrt{100-n}} \right)$$

where $$n=1,2,...99$$. As a result,

$$N= \frac{1}{\sqrt2}(N+D), \>\>\>\>\>D= \frac{1}{\sqrt2}(N-D)$$ Take the ratio,

$$\frac ND=\frac{N+D}{N-D}$$

or,

$$\left(\frac ND\right)^2 - 2\frac ND -1 =0$$

Solve to obtain,

$$\frac ND = 1+\sqrt2$$

• Nice solution (+1). Where did you find those denesting formulas? Oct 22, 2019 at 16:28
• @Axion004 - It is not hard to derive it, or see the link en.wikipedia.org/wiki/Nested_radical Oct 22, 2019 at 16:31

Rewrite the sum as $$\frac{\sum_{n=1}^{99} \sqrt{10+\sqrt{n}}}{\sum_{n=1}^{99} \sqrt{10-\sqrt{n}}} = \frac{a}{b}$$ then let $$\Delta_n = \sqrt{10+\sqrt{n}} - \sqrt{10-\sqrt{n}}$$ By squaring $$\Delta_n$$ and simplifying we get $$\Delta_n = \sqrt{2} \sqrt{10 - \sqrt{100-n}}$$ Now summing all those $$\Delta_n$$s can be done by letting the index $$n$$ run from $$1$$ to $$99$$ and replacing $$\sqrt{100-n}$$ with $$\sqrt{n}$$ since both cases will yield the same summands. Hence $$\sum_{n=1}^{99} \Delta_n = \sqrt{2} \sum_{n=1}^{99}\sqrt{10 - \sqrt{n}} = \sqrt{2} \cdot b$$ thus $$\frac{a-b}{b} = \frac{a}{b} - 1 = \sqrt{2} \implies \frac{a}{b} = \sqrt{2} + 1$$

• Thank you very much again. Jul 16, 2020 at 22:06

I'd like to add one more answer "by color-coding". First, observe that $$\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}} = \sqrt{2}\sqrt{n-\sqrt{n^2-k}}.$$ Now the problem. Designate $$x = \frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}},$$ and calculate $$x-1 = \frac{\overbrace{\sqrt{10+\sqrt{1}}-\sqrt{10-\sqrt{1}}}^{\color{red}{\sqrt{2}\sqrt{10-\sqrt{99}}}} + \cdots + \overbrace{\sqrt{10+\sqrt{99}}-\sqrt{10-\sqrt{99}}}^{\color{blue}{\sqrt{2}\sqrt{10-\sqrt{1}}}} }{\color{blue}{\sqrt{10-\sqrt{1}}}+\cdots+\color{red}{\sqrt{10-\sqrt{99}}}}.$$ Everything simplifies to $$\sqrt{2}$$, thus $$x = 1 + \sqrt{2}$$

• @Jyrki Lahtonen Yes, you are perfectly right. I did a typo due to copy-paste. Many thanks! Oct 22, 2019 at 20:52