What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series?

For a Balmer series:

$\frac{1}{{\lambda }_B}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right] ....\left(i\right)$

where $n=3,4,........$

By putting $n=\infty$ in equation (i), we obtain the series limit of the Balmer series. This is the shortest wavelength of the Balmer series. 

or ${\lambda }_B=\frac{4}{R} ....\left(ii\right)$

For a Lyman series

$\frac{1}{{\lambda }_L}=R\left[\frac{1}{1^2}-\frac{1}{n^2}\right] \left(iii\right)$

where $n=2,3,4,...$

By putting $n=\infty$ in equation (iii), we obtain the series limit of the Lyman series. This is the shortest wavelength of the Lyman series.

or   ${\lambda }_L=\frac{1}{R} ....\left(iv\right)$

Dividing (ii) by (iv), we get,  $\frac{{\lambda }_B}{{\lambda }_L}=\frac{4}{1}$