The shortest wavelength (series limit) corresponds to the transition from $n_2 = \infty$. For the Balmer series, the transition is to $n_1 = 2$ . Using the Rydberg formula $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, we get $\frac{1}{\lambda} = R \left(\frac{1}{2^2} - 0\right) = \frac{R}{4}$, so $\lambda = \frac{4}{R}$. For the Brackett series, the transition is to $n_1 = 4$ . The shortest wavelength $\lambda'$ is given by $\frac{1}{\lambda'} = R \left(\frac{1}{4^2} - 0\right) = \frac{R}{16}$, so $\lambda' = \frac{16}{R}$. Comparing the two, $\lambda' = 4 \left(\frac{4}{R}\right) = 4\lambda$.