Calculate Square of Sum ∑1..9999

[Albert1]

\(\displaystyle \left(\sum_{n=1}^{9999}\frac{\sqrt{100+\sqrt{n}}}{\sqrt{100-\sqrt{n}}}+\sum_{n=1}^{9999}\frac{\sqrt{100-\sqrt{n}}}{\sqrt{100+\sqrt{n}}}\right)^2\)

 

[Prove It]

$(\sum_{n=1}^{9999}\dfrac{\sqrt{100+\sqrt n}}{\sqrt{100-\sqrt n}}
+\sum_{n=1}^{9999}\dfrac{\sqrt{100-\sqrt n}}{\sqrt{100+\sqrt n}})^2$

I would try to simplify these sums by multiplying the top and bottom of each by the bottom's conjugate.

 

[Albert1]

$(\sum_{n=1}^{9999}\dfrac{\sqrt{100+\sqrt n}}{\sqrt{100-\sqrt n}}
+\sum_{n=1}^{9999}\dfrac{\sqrt{100-\sqrt n}}{\sqrt{100+\sqrt n}})^2$

sorry it should be:

\(\displaystyle \left(\frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}
+\frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}\right)^2\)

 

[Opalg]

\(\displaystyle \left(\frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}
+\frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}\right)^2\)

8
[sp]Generalisation. This is a particular case ($N=100$) of this problem:

Find \(\displaystyle \biggl(f(N) + \frac1{f(N)}\biggr)^{\!2}\), where \(\displaystyle f(N) = \frac{\sum\limits_{n=1}^{N^2-1}\sqrt{N+\sqrt n}}{\sum\limits_{n=1}^{N^2-1}\sqrt{N-\sqrt n}}.\)
​

Experimentation. To see which way the wind is blowing, I got out my trusty pocket calculator and found a numerical expression for $f(2)$: $$f(2) = \frac{\sqrt{2 + \sqrt1} + \sqrt{2 + \sqrt2} + \sqrt{2 + \sqrt3}} {\sqrt{2 - \sqrt1} + \sqrt{2 - \sqrt2} + \sqrt{2 - \sqrt3}} \approx \frac{1.732 + 1.848 + 1.932}{1.000 + 0.765 + 0.518} = \frac{5.512}{2.283} \approx 2.414.$$ I recognised that answer as being suspiciously close to $1+\sqrt2$, but I could not see either how to prove that or how to generalise it for higher values of $N$. So I did the same calculation for $N=3$, and found to my surprise that answer again looked like $1+\sqrt2$. This made me wonder what the denominator of the fraction for $f(2)$ would look like if I multiplied it by $\sqrt2$, and that led to the following table: $$ \begin{array}{r|cccl} \text{row 1} & 1.732 & 1.848 & 1.932 & \text{(numerator)} \\ \text{row 2} & 1.000 & 0.765 & 0.518 & \text{(denominator)} \\ \text{row 3} & 1.414 & 1.082 & 0.732 & \text{(row 2 times $\sqrt2$)} \\ \text{row 4} & 0.732 & 1.082 & 1.414 & \text{(row 3 reversed)} \\ \text{row 5} & 1.732 & 1.848 & 1.932 & \text{(row 2 plus row 4)} \end{array}$$ Row 5 is the same as row 1! That gave me the idea for proving that $f(N) = 1 + \sqrt2$ for all $N$.

Solution. For $1\leqslant n< N$, $$\begin{aligned} 2\bigl(N - \sqrt{N^2-n}\bigr) &= 2N - 2\sqrt{N^2-n} \\ &= (N + \sqrt n) - 2 \sqrt{(N+\sqrt n)(N - \sqrt n)} + (N - \sqrt n) \\ &= \bigl(\sqrt{N + \sqrt n} - \sqrt{N - \sqrt n}\bigr)^2. \end{aligned}$$ Take the positive square root of both sides, to get $\sqrt2\sqrt{N - \sqrt{N^2-n}} = \sqrt{N + \sqrt n} - \sqrt{N - \sqrt n}$ and therefore $\sqrt{N - \sqrt n} + \sqrt2\sqrt{N - \sqrt{N^2-n}} = \sqrt{N + \sqrt n}.$ Sum that from $n=1$ to $N^2-1$: $$ \sum_{n=1}^{N^2-1}\sqrt{N - \sqrt n}\: + \sum_{n=1}^{N^2-1}\sqrt2\sqrt{N - \sqrt{N^2-n}} = \sum_{n=1}^{N^2-1}\sqrt{N + \sqrt n}.$$ Now reverse the order of summation for the second sum on the left side, by replacing $n$ by $N^2-n$: $$ \sum_{n=1}^{N^2-1}\sqrt{N - \sqrt n}\: + \sum_{n=1}^{N^2-1}\sqrt2\sqrt{N - \sqrt n} = \sum_{n=1}^{N^2-1}\sqrt{N + \sqrt n}.$$ The left side is then \(\displaystyle \bigl(1+\sqrt2\bigr)\sqrt{N - \sqrt n}\), from which it follows that $f(N) = 1+\sqrt2.$

Conclusion. \(\displaystyle \biggl(f(N) + \frac1{f(N)}\biggr)^{\!2} = \bigl((1+\sqrt2) + (1+\sqrt2)^{-1}\bigr)^2 =\bigl((1+\sqrt2) + (\sqrt2 - 1)\bigr)^2 = \bigl(2\sqrt2\bigr)^2 = 8.\)[/sp]

 

[anemone]

Opalg's has given one so detail solution and my hat is off to him!

My solution that is essentially quite the same as Opalg's method:

Spoiler

If we let

$a=\sqrt{100+\sqrt n}-\sqrt{100-\sqrt n}$

Then squaring, rearranging and taking square root of it we have

$a^2=100+\sqrt n+100-\sqrt n-2(\sqrt{100+\sqrt n})(\sqrt{100-\sqrt n})$

$a^2=200--2\sqrt{100^2-n})$

$a^2=2(100-\sqrt{100^2-n})$

$a=\sqrt{2}\sqrt{(100-\sqrt{100^2-n})}$

$\therefore \sqrt{2}\sqrt{(100-\sqrt{100^2-n})}=\sqrt{100+\sqrt n}-\sqrt{100-\sqrt n}$

Notice that

$\begin{align*}\displaystyle \sum\limits_{n=1}^{9999}\sqrt{2}\sqrt{(100-\sqrt{100^2-n})}&=\sqrt{2}\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{100^2-n})}\\&=\sqrt{2}\left(\sqrt{100-\sqrt{9999}}+\sqrt{100-\sqrt{9998}}+\cdots +\sqrt{100-\sqrt{2}}+\sqrt{100-\sqrt{1}}\right)\\&=\sqrt{2}\left(\sqrt{100-\sqrt{1}}+\sqrt{100-\sqrt{2}}+\cdots +\sqrt{100-\sqrt{9998}}+\sqrt{100-\sqrt{9999}}\right)\\&=\sqrt{2}\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}\end{align*}$

So we have
$\displaystyle \sum\limits_{n=1}^{9999} \sqrt{2}\sqrt{(100-\sqrt{100^2-n})}= \sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}- \sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}$

$\displaystyle\sqrt{2}\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}=\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}- \sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}$

$\displaystyle\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}(\sqrt{2}+1)=\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}$

Replacing this into the first fraction inside the square of the expression gives

$\begin{align*}\displaystyle \frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}&=\frac{\sum\limits_{n=1}^{9999}\sqrt{(100-\sqrt{n})}(\sqrt{2}+1)}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}\\&=\sqrt{2}+1\end{align*}$

Hence,

$\begin{align*}\displaystyle \frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}&=\dfrac{1}{\sqrt{2}+1}\\&=\sqrt{2}-1 \end{align*}$

Last, we see that

$\displaystyle \left(\frac{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}
+\frac{\sum\limits_{n=1}^{9999}\sqrt{100-\sqrt n}}{\sum\limits_{n=1}^{9999}\sqrt{100+\sqrt n}}\right)^2=(\sqrt{2}+1+\sqrt{2}-1)^2=8$