In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy
This is an excellent question! Just to elaborate upon what @porphyrn and @M. Farooq have said, imagine that the universe is entirely empty except for one sodium atom and one photon (particle of light) that has a wavelength of $\pu{242 nm}$. If the photon bumps into the sodium atom, it might just bounce off: this is called scattering. But another big possibility is that the photon will be absorbed (i.e., ‘taken in’, as it were) by sodium’s 3s valence electron. The energy of a $\pu{242 nm}$ photon is $E_{\pu{242 nm}} = h\nu = \frac{hc}\lambda$, where $h = \pu{6.626 x 10^{-34} Js}$ and $c = \pu{2.9979E8 m/s} = \pu{2.9979E17 nm/s}$. This energy is $\pu{8.21E-19 J}$.

Then the sodium’s 3s electron would have just enough energy to escape, i.e., ‘break free’, from the sodium atom. The result would be a sodium ion, $\ce{Na+}$, and an electron, $\ce {e-}$, that were free from one another: the sodium atom has been ionized, i.e., has become an ion, so the energy of the $\pu{242 nm}$ photon is the ionization energy of the 3s electron in sodium. Using the $h$ and $c$ values, the required ionization energy, for one sodium atom, is $\pu{8.21 x 10^{-19} J}$. For a mole of sodium atoms, multiply by Avogadro’s number, resulting in $\pu{494 kJ/mol}$.

So where does the infinity enter the picture? Well, a positively charge sodium ion and the electron that escaped from it attract: opposite charges attract. The Coulombic force of attraction is an inverse square law, similar to gravity, so total absolute freedom would mean that the electron was infinitely far away from the sodium ion and had zero kinetic energy, i.e., zero energy of motion. But, as a practical matter, the separation can be relatively small, not infinity! As a practical matter, it is known experimentally that an isolated sodium atom, doing its thing in a vacuum, can be ionized by shining $\pu{242 nm}$ light on it.

In your example, suppose a sodium ion, in vacuum, happened to recombine with a free electron. Assume the sodium ion and electron are moving very slowly relatively, so there is negligible excess kinetic energy: we don't want the electron to 'harpoon' the poor sodium ion! The electron could drop down into any empty orbital of the sodium ion. Potentially, that is an infinite number of possibilities, hence M. Farooq's answer (and sweet Grotian diagram). The result would be release of a photon with wavelength longer (numerically 'larger') than $\pu{242 nm}$. This excited (it has more energy than a 'ground state') sodium atom must eventually drop to the 'ground state' (the only state where it is indefinitely stable) by emitting one or more additional photons or perhaps getting 'mugged' of its excess energy, via collision with something else. This is collisional deactivation.

But what if the electron dropped all the way down, in one fell swoop, to the 3s orbital? Then the resulting sodium atom would be in its 'ground state', i.e., lowest energy (and only indefinitely stable) state, and the emitted photon would be at $\pu{242 nm}$.