Solution:

The shortest wavelength in a series corresponds to transition from \(n_2 = \infty\) to \(n_1\). For Lyman series, \(\lambda_L = \frac{1}{R}\). For Balmer series, \(\lambda_B = \frac{4}{R}\). Therefore, the ratio \(\frac{\lambda_L}{\lambda_B} = \frac{1}{4}\).