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

For any hydrogen‐like spectrum the Rydberg formula is
$$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),\qquad n_2 \gt n_1,\; R=1.097\times10^{7}\,\text{m}^{-1}$$

Balmer series has the lower level $$n_1=2$$.
Shortest wavelength (highest energy) ⇒ $$n_2=\infty$$:
$$\frac{1}{\lambda_{\min}}=R\left(\frac{1}{2^2}-0\right)=\frac{R}{4}\;\Rightarrow\;\lambda_{\min}=\frac{4}{R}$$
Longest wavelength (lowest energy) ⇒ $$n_2=3$$:
$$\frac{1}{\lambda_{\max}}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5R}{36}$$
$$\lambda_{\max}=\frac{36}{5R}$$
Hence
$$\frac{\lambda_{\max}}{\lambda_{\min}}=\frac{36/(5R)}{4/R}=\frac{36}{20}=\frac{9}{5}$$
Statement A is correct.

Numerical limits of the two series:

Balmer: $$\lambda_{\text{range}}=365\text{ nm to }656\text{ nm}$$
Paschen ($$n_1=3$$):

Shortest: $$n_2=\infty\;\Rightarrow\;\lambda_{\min}= \frac{9}{R}=821\text{ nm}$$
Longest: $$n_2=4\;\Rightarrow\;\frac{1}{\lambda_{\max}}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{144}$$
$$\lambda_{\max}=\frac{144}{7R}=1875\text{ nm}$$

Balmer ends at 656 nm, Paschen begins at 821 nm; there is a clear gap, so no overlap.
Statement B is incorrect.

Lyman series ($$n_1=1$$): $$\dfrac{1}{\lambda_m}=R\left(1-\dfrac{1}{m^2}\right)$$ with $$m=2,3,\ldots$$
Let $$\lambda_0$$ be the shortest wavelength ( $$n_2=\infty$$ ) : $$\lambda_0=\dfrac{1}{R}$$.

Actual relation:
$$\lambda_m=\frac{1}{R\left(1-\dfrac{1}{m^2}\right)}=\lambda_0\;\frac{1}{1-\dfrac{1}{m^2}}=\lambda_0\;\frac{m^2}{m^2-1}$$
This is not equal to $$\left(1+\frac{1}{m^2}\right)\lambda_0$$.
Statement C is incorrect.

Lyman wavelengths run from 91 nm (shortest) to 122 nm (longest).
Balmer wavelengths run from 365 nm to 656 nm.
These intervals are far apart, hence they do not overlap.
Statement D is correct.

Therefore the correct statements are:
Option A (ratio $$9/5$$ in Balmer series) and Option D (no overlap of Lyman and Balmer series).

Option A, Option D