Which of the following statement(s) is (are) correct about the spectrum of hydrogen atom?

When transition is from any level to n = 2, then it belongs to Balmer series for which longest wavelength is corresponding to least energy difference i.e. for transition from n = 3 to n = 2 and shortest wavelength is corresponding to maximum energy difference i.e. for transition  $n=\infty ton=2$ . The wavelength can be calculated by Rydberg’s formula given as
$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ 
$\Rightarrow \frac{{\lambda }_{longest}}{{\lambda }_{shorest}}=\frac{[\frac{1}{2^2}-\frac{1}{\propto }]}{[\frac{1}{2^2}-\frac{1}{3^2}]}=\frac{9}{5}$ 
Hence option (A) is correct.
The longest wavelength of Balmer series is given as 
${\lambda }_{longest}=\frac{36}{5R}$ 
The shortest wavelength of paschen series is given as 
${\lambda }_{shorest}=\frac{9}{R}$ 
These do not overlap hence option (B) is NOT correct.
Wavelength of Lyman series from m^th orbit is given as 
 $\frac{1}{\lambda }=R[\frac{1}{1^2}-\frac{1}{m^2}]$
Shortest wavelength of Lyman series is corresponding to the transition from $n=\infty ton=1$  which is given as

  $\frac{1}{{\lambda }_0}=R$
 $\Rightarrow \frac{1}{\lambda }-\frac{1}{{\lambda }_0}[1-\frac{1}{m^2}]$
 $\Rightarrow \lambda \frac{{\lambda }_0}{1-\frac{1}{m^2}}$
Hence option (C) is NOT correct.
The longest wavelength of Lyman series is given as 
 ${\lambda }_{longest}=\frac{4}{3R}$
The shortest wavelength of Balmer series is given as
 ${\lambda }_{shorest}=\frac{4}{R}$
These wavelength do not overlap hence option (D) is correct.