Dividing by $\sqrt n$ Why is the following equality true? 
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
 A: Alternate form:
$$\frac{\frac{\sqrt{n}}{ \sqrt{n}}}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}}{\sqrt{n}}}=\frac{1}{\sqrt{\frac{n+\sqrt{n+\sqrt{n}}}{n}}}=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{n}}}=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{\sqrt{n} \sqrt{n}}}}=\frac{1}{\sqrt{1+ \frac{1}{\sqrt{n}} \frac{\sqrt{n+\sqrt{n}}}{ \sqrt{n}}}}=\frac{1}{\sqrt{1+\frac{1}{\sqrt{n}} \sqrt{\frac{n+\sqrt{n}}{n}}}}=\frac{1}{\sqrt{1+\frac{1}{\sqrt{n}} \sqrt{1+\frac{1}{\sqrt{n}}}}} $$
A: Suppose that $n>0$.
$$\frac{\sqrt{n}}{ \sqrt{n+\sqrt{n+\sqrt{n}}} } = \frac{1}{ \frac{1}{\sqrt{n}} \left( \sqrt{n+\sqrt{n+\sqrt{n}}} \right) } = \frac{1}{ \sqrt{1+ \frac{1}{n} \left( \sqrt{n + \sqrt{n}} \right) } } = \frac{1}{\sqrt{1+ \sqrt{ \frac{1}{n} + \frac{1}{n^2} \sqrt{n} } } } = \frac{1}{\sqrt{1+ \sqrt{ \frac{1}{n} + \sqrt{ \frac{1}{n^3}} } } } $$.
A: I find this easier to visualize if I write this problem in terms of powers:
$$\dfrac{n^{1/2}}{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}\text{.} $$
Division by $n^{1/2}$ for both the numerator and denominator turns the numerator into $1$ and the denominator into
$$\begin{align*}
\dfrac{\left[n + \left(n + n^{1/2}\right)^{1/2}\right]^{1/2}}{n^{1/2}} &= \left[\dfrac{n + \left(n + n^{1/2}\right)^{1/2}}{n}\right]^{1/2}\quad \text{ since }\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^{c} \\
&= \left[\dfrac{n}{n} + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{n}\right]^{1/2} \quad \text{ since }\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \\
&= \left[1 + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{n}\right]^{1/2} \\
&= \left[1 + \dfrac{\left(n+n^{1/2}\right)^{1/2}}{\left(n^{2}\right)^{1/2}}\right]^{1/2} \\
&= \left[1 + \left(\dfrac{n+n^{1/2}}{n^{2}}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a^c}{b^c} = \left(\dfrac{a}{b}\right)^{c} \\
&= \left[1 + \left(\dfrac{n}{n^2}+\dfrac{n^{1/2}}{n^2}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \\
&= \left[1 + \left(\dfrac{1}{n}+\dfrac{1}{n^{3/2}}\right)^{1/2}\right]^{1/2} \text{ since }\dfrac{a^b}{a^c} = a^{b-c} \\
&= \left\{1 + \left[\dfrac{1}{n}+\dfrac{1^{1/2}}{(n^{3})^{1/2}}\right]^{1/2}\right\}^{1/2} \\
&= \left\{1 + \left[\dfrac{1}{n}+\left(\dfrac{1}{n^3}\right)^{1/2}\right]^{1/2}\right\}^{1/2} \\
&= \sqrt{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}}}}\text{.}
\end{align*}$$
A: Well this is calculus. As you said yourself, you divide the numerator and the denominator by $\sqrt{n}$
for the numerator it's easy, as $\frac{\sqrt{n}}{\sqrt{n}}=1$.
for the denominator you get \begin{equation}
\frac{1}{\sqrt{n}}\sqrt{n+{\sqrt{n+{\sqrt{n}}}}}=\sqrt{1+\frac{\sqrt{n+\sqrt{n}}}{n}}
\end{equation}
Keep in mind, if you want to pull a factor under a square root, you have to square it!
the above can be rewritten as \begin{equation}
\sqrt{1+\sqrt{\frac{n}{n^2}+\frac{\sqrt{n}}{n^2}}}= \sqrt{1+\sqrt{\frac{1}{n}+\sqrt{\frac{n}{n^4}}}}=\sqrt{1+\sqrt{\frac{1}{n}+\sqrt{\frac{1}{n^3}}}}
\end{equation}
And you're done
A: Slightly modifying the answer from @evinda:
$\begin{align}
\frac{\frac{\sqrt{n}}{ \sqrt{n}}}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}}{\sqrt{n}}}&=\frac{1}{\sqrt{\frac{n+\sqrt{n+\sqrt{n}}}{n}}} \\
&=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{n}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac{n+\sqrt{n}}{n^2}}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac1n+\frac{\sqrt{n}}{n^2}}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac1n+\sqrt{\frac{n}{n^4}}}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac1n+\sqrt{\frac{1}{n^3}}}}}
\end{align}$
