Voltage drops in series - less work to do the same job

So I have been trying to explain electricity to younger siblings and it surprised me how little I know of what actually goes on.

My question comes in the form of a consideration:

Consider a circuit with a battery and a bulb. The battery has a voltage $$V$$. The potential difference across the bulb is then also $$V$$. This is the work that is done to move a unit charge across the bulb. Now consider two identical bulbs in series in the same circuit. It now only takes $$\frac{V}{2}$$ of work to move a unit charge across each bulb.

I may be thinking about this entirely wrong, but if I take the number of bulbs to infinity, then a current will still flow (?) and the amount of work to move a unit charge across the bulb tends to $$0$$? As such, it takes increasingly less work to do the same job.

It seems to allude to something like a perpetual motion machine. So why bother spending any work doing it at all?

Of course, I realise I must have a misconception somewhere, but it's just where that is that I am not too sure.

I take the number of bulbs to infinity, then a current will still flow (?)

Each bulb is a resistor.

If you take the number of bulbs to infinity, then the equivalent resistance of the series combination of bulbs goes to infinity. Therefore the current through the bulbs goes to zero.

... and the amount of work to move a unit charge across the bulb tends to 0? As such, it takes increasingly less work to do the same job.

The amount of work done per unit charge is going to zero, but the rate that charge is moving through the system also goes to zero. In fact it goes to zero even faster than the work done per unit charge goes to zero, with the result that the amount of power transferred from the battery to the series combination of bulbs also goes to zero, and therefore the amount of light produced also goes to zero.

if I take the number of bulbs to infinity, then a current will still flow

Stipulate that the two bulbs are identical and that they act as ideal resistors (constant resistance) with resistance $$R$$. With just one bulb, the current $$I$$ through is just

$$I_1 = \frac{V}{R}$$

With two series connected bulbs, the current is

$$I_2 = \frac{V}{2R} = \frac{I_1}{2}$$

With $$N$$ bulbs,

$$I_N = \frac{I_1}{N}$$

and it follows that

$$I_\infty = 0$$