How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$
How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?
In general, it holds that $$\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}=n-\frac{1}{8n^2}+O\left(\frac1{n^3}\right)$$ and that $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac1{n^3}\right)\,$$ for all $n\geq 1$. Hence, their difference is $$1-\frac1{4n^2}+O\left(\frac1{n^3}\right)\,.$$ In particular, for $n=5$, the difference should be about $1-\dfrac1{100}=0.99$. This is quite close to the actual value of $0.9872649...$.
From $\sqrt{1+x}=1+\frac{1}{2}x+O\left(x^2\right)$, we have $\sqrt{1-\frac1n}=1-\frac1{2n}+O\left(\frac{1}{n^2}\right)$. This means $$\sqrt{n(n-1)}=n\,\sqrt{1-\frac{1}{n}}=n\,\Biggl(1-\frac1{2n}+O\left(\frac{1}{n^2}\right)\Biggr)=n-\frac{1}{2}+O\left(\frac{1}{n}\right)\,,$$ which can also be written as $$\sqrt{n(n-1)}=(n-1)+\frac{1}{2}+O\left(\frac{1}{n}\right)\,.$$ Ergo, $$\begin{align}\sqrt{n(n-1)+\sqrt{n(n-1)}}&=\sqrt{n^2-\frac{1}{2}+O\left(\frac{1}{n}\right)}&=&n\,\sqrt{1-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)}\\&=n\,\Biggl(1-\frac{1}{4n^2}+O\left(\frac{1}{n^3}\right)\Biggr)&=&n-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\,.\end{align}$$ Similarly, $$\sqrt{n(n-1)-\sqrt{n(n-1)}}=(n-1)-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\,.$$ Hence, $$\begin{align}\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}&=\sqrt{n^2-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)}\\&=n\,\sqrt{1-\frac{1}{4n^3}+O\left(\frac{1}{n^4}\right)}\\&=n\,\Biggl(1-\frac{1}{8n^3}+O\left(\frac{1}{n^4}\right)\Biggr)\\&=n-\frac{1}{8n^3}+O\left(\frac{1}{n^3}\right)\,.\end{align}$$ Likewise, $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac{1}{n^3}\right)\,.$$ In fact, we can prove by induction on $k$ that $$f^+_k\big(n(n-1)\big)=n-\frac{1}{2^kn^{k-1}}+O\left(\frac{1}{n^k}\right)$$ and $$f^-_k\big(n(n-1)\big)=(n-1)-\frac{(-1)^k}{2^kn^{k-1}}+O\left(\frac{1}{n^k}\right)\,,$$ where $f^+_k(x):=\sqrt{x+f^+_{k-1}(x)}$ with $f^+_0(x):=0$ and $f^-_k(x):=\sqrt{x-f^-_{k-1}(x)}$ with $f^-_0(x):=0$ for all $x\geq 0$ and for each $k=1,2,3,\ldots$.
Even without a calculator, you can do the numerics fairly easily. 20 is about halfway between 16 and 25, so $\sqrt{20} \approx 4.5$. So $20\pm\sqrt{20}$ is about 15.5 and 24.5, respectively. These in turn have square roots of about (a little less than) 4 and 5. This leaves $\sqrt{25}−\sqrt{16}\approx 1$. The "little less than" might contribute significant error on its own, but since it's similar sizes we can expect it to largely cancel out: $\sqrt{25-\epsilon} - \sqrt{16-\epsilon} \approx \sqrt{25} - \sqrt{16}$, since we can expect it shift each root over by a similar amount. Here $\epsilon = 1/2$.
If we consider an infinite chain.
Suppose $x = \sqrt{20 +\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{\cdots}}}}}$
$x = \sqrt{20 +x}\\ x^2 = 20 + x\\ x^2 - x - 20 = 0\\ (x-5)(x+4) = 0$
$x$ must be greater than $0, x = 5$
and $y = \sqrt{20 -\sqrt{20-\sqrt{20-\sqrt{20-\sqrt{\cdots}}}}}$
$y = \sqrt{20 - y}\\ y^2 + y - 20=0\\ y = 4$
$x-y = 1$
As we add more terms under those square roots $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}}$ converges toward 1.
Use the classical approximation: $$\sqrt{a^2 + b} \approx a + \frac{b}{2a}$$ With $a = \sqrt{20}$ and $b = \sqrt{20 + \sqrt{20}}$ we have $$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{\sqrt{20 + \sqrt{20}}}{2\sqrt{20}} = \sqrt{20} + \frac{\sqrt{400 + 20\sqrt{20}}}{40} $$ Now use the same classical approximation again, this time working with the numerator of the second term. This time with $a=20$ and $b=20\sqrt{20}$ we get $$\sqrt{400 + 20\sqrt{20}} \approx 20 + \frac{20\sqrt{20}}{40} = 20 + \frac{\sqrt{20}}{2}$$ Combining these, we've got: $$\sqrt{20 + \sqrt{20 + \sqrt{20}}} \approx \sqrt{20} + \frac{20 + \frac{\sqrt{20}}{2}}{40} = \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80}$$
Using similar methods, we get $$\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80}$$
Finally, subtracting one from the other we end up with $$\sqrt{20 + \sqrt{20 + \sqrt{20}}} -\sqrt{20 - \sqrt{20 - \sqrt{20}}} \approx \left( \sqrt{20} + \frac{1}{2} + \frac{\sqrt{20}}{80} \right) - \left( \sqrt{20} - \frac{1}{2} + \frac{\sqrt{20}}{80} \right)$$ and in this last expression everything cancels out except for $$\frac{1}{2} - \left(-\frac{1}{2} \right) = 1$$