Shortest wavelength in Balmer series when transition of $e^{-}$ from $\infty$ to $\mathrm{n}=2$

$$\because \frac{1}{\lambda}=\mathrm{Rz}^{2}\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right]$$

$$\frac{1}{\lambda}=\frac{R}{4}$$ .... (1)

Shortest wavelength is Bracket series when transition of $e^{-}$ from $\infty$ to $\mathrm{n}=4$

$$\frac{1}{\lambda^{\prime}}=\mathrm{R}(1)^{2}\left[\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right] \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{\mathrm{R}}{16}$$ ..... (2)

Eq. (1) / Eq. (2)

$$\frac{\lambda^{\prime}}{\lambda}=\frac{\mathrm{R}}{4} \times \frac{16}{\mathrm{R}} \Rightarrow \lambda^{\prime}=4 \lambda$$