Show that the first few frequencies of light that is emitted when electrons fall to the $n^{\text{th}}$ level from levels higher than $n$,are approximate harmonics (i.e.,in the ratio $1:2:3...$) when $n >> 1$.

(N/A) In an atom of an element with atomic number $Z$,when an electron makes a transition from the $(n+x)^{\text{th}}$ orbit (where $x=1, 2, 3, \dots$) to the $n^{\text{th}}$ orbit,the frequency $f$ of the emitted radiation is given by the Rydberg formula:
$\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n^{2}} - \frac{1}{(n+x)^{2}} \right)$
Since $f = \frac{c}{\lambda}$,we have:
$f = c R Z^{2} \left( \frac{(n+x)^{2} - n^{2}}{n^{2}(n+x)^{2}} \right)$
$f = c R Z^{2} \left( \frac{n^{2} + 2nx + x^{2} - n^{2}}{n^{2}(n+x)^{2}} \right)$
$f = c R Z^{2} \left( \frac{2nx + x^{2}}{n^{2}(n+x)^{2}} \right)$
Given $n >> 1$ and $x$ is small $(x=1, 2, 3, \dots)$,we can approximate $(n+x) \approx n$:
$f \approx c R Z^{2} \left( \frac{2nx}{n^{2} \cdot n^{2}} \right)$
$f \approx \left( \frac{2 R c Z^{2}}{n^{3}} \right) x$
Since $\frac{2 R c Z^{2}}{n^{3}}$ is constant for a fixed $n$,we have $f \propto x$.
Therefore,the ratio of frequencies for $x=1, 2, 3$ is $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$.