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?
 A: 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 $\epsilon = 1/2$.
A: 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.
A: 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
$$\frac{1}{2} - \left(-\frac{1}{2} \right) = 1$$
A: Repeatedly using
$\sqrt{1+x}
\approx 1+x/2$,
$\begin{array}\\
d(a)
&=\sqrt{a+\sqrt{a+\sqrt{a}}}-\sqrt{a-\sqrt{a-\sqrt{a}}}\\
&=(\sqrt{a+\sqrt{a+\sqrt{a}}}-\sqrt{a-\sqrt{a-\sqrt{a}}})\dfrac{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\
&=\dfrac{(a+\sqrt{a+\sqrt{a}})-(a-\sqrt{a-\sqrt{a}})}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\
&=\dfrac{\sqrt{a+\sqrt{a}}+\sqrt{a-\sqrt{a}}}{\sqrt{a+\sqrt{a+\sqrt{a}}}+\sqrt{a-\sqrt{a-\sqrt{a}}}}\\
&=\dfrac{\sqrt{a}(\sqrt{1+1/\sqrt{a}}+\sqrt{1-1/\sqrt{a}})}{\sqrt{a}\sqrt{1+(1/a)\sqrt{a+\sqrt{a}}}+\sqrt{a}\sqrt{1-(1/a)\sqrt{a-\sqrt{a}}}}\\
&=\dfrac{\sqrt{1+1/\sqrt{a}}+\sqrt{1-1/\sqrt{a}}}{\sqrt{1+\sqrt{1/a+1/a^{3/2}}}+\sqrt{1-\sqrt{1/a-1/a^{3/2}}}}\\
&\approx\dfrac{1+1/(2\sqrt{a})+1-1/(2\sqrt{a})} {1+(1/2)\sqrt{1/a+1/a^{3/2}}+1-(1/2)\sqrt{1/a-1/a^{3/2}}}\\
&=\dfrac{2} {2+(1/(2\sqrt{a}))\sqrt{1+1/a}-(1/(2\sqrt{a}))\sqrt{1-1/a}}\\
&\approx\dfrac{2} {2+(1/(2\sqrt{a}))(1+1/(2a)-(1/(2\sqrt{a}))(1-1/(2a)}\\
&\approx\dfrac{2} {2+(1/(2\sqrt{a}))(1/(2a))}\\
&=\dfrac{1} {1+(1/(8a^{3/2}))}\\
&\approx 1-(1/(8a^{3/2}))\\
\end{array}
$
A: By binomial formulas and cancellation, you get that
\begin{align}
...&=\frac{\sqrt{20+\sqrt{20}}+\sqrt{20-\sqrt{20}}}{\sqrt{20+\sqrt{20+\sqrt{20}}}+\sqrt{20-\sqrt{20-\sqrt{20}}}}
\\&=\frac{
  \sqrt{5+\sqrt{\frac54}}+\sqrt{5-\sqrt{\frac54}}
  }{
  \sqrt{5+\sqrt{\frac54+\sqrt{\frac5{64}}}}+\sqrt{5-\sqrt{\frac54-\sqrt{\frac5{64}}}}
  }
\end{align}
As one can see, the denominator is a small perturbation of the numerator, so that the quotient will be close to $1$.
A: 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 $k=1,2,3,\ldots$.

A: As in @mweiss 's answer we use repeatedly the approximation
$$\sqrt{a^2+b}\approx a+{b\over 2a}\qquad(|b|\ll a^2)\ .$$In this way we obtain on the one hand
$$\eqalign{
\sqrt{20}&=\sqrt{25-5}\approx 5-{1\over2},\quad
20+\sqrt{20}\approx25-{1\over2},\cr
\sqrt{20+\sqrt{20}}&\approx5-{1\over20},\quad 20+\sqrt{20+\sqrt{20}}\approx 25-{1\over20},\cr
\sqrt{20+\sqrt{20+\sqrt{20}}}&\approx5-{1\over200},
\cr}$$
and on the other hand
$$\eqalign{
\sqrt{20}&=\sqrt{16+4}\approx 4+{1\over2},\quad
20-\sqrt{20}\approx16-{1\over2},\cr
\sqrt{20-\sqrt{20}}&\approx4-{1\over16},\quad 20-\sqrt{20-\sqrt{20}}\approx 16+{1\over16},\cr
\sqrt{20-\sqrt{20-\sqrt{20}}}&\approx4+{1\over128}.
\cr}$$
It follos that the quantity $Q$ in question is approximatively given by
$$Q\approx1-{1\over200}-{1\over128}\approx0.9872.$$
