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 $k=1,2,3,\ldots$.

