The 40 watt bulb will be brighter. For discussion, let's assume that light bulbs are strictly resisters. They're not, light bulbs' resistance increases with the temperature of the filament. But for discussion: A 40 W bulb (120V) is 360 ohms, a 60 W bulb is 240 ohms. Now, we have both in series fed by 120V. Total resistance is 600 ohms. The 40 W bulb will see 3/5 of 120V, and the 60W will see 2/5 of 120V (think resistive voltage divider). The current will be the same thru both bulbs, so as the 40 W bulb sees more voltage across it, will dissipate more power. Thus glow brighter.

Now consider that the resistance of a light bulb is lower when cold, the 60W will see even less voltage across it. Which means the 40W will see more voltage. And the 40W bulb's resistance will go even higher as it gets hotter. And thus get a bit brighter. This resistance increase is not linear. But the answer to the question is that the 40w bulb will be brighter.

I've applied this fact on Christmas light strings, some with blinkers. Series/parallel connection. See http://home.netcom.com/~wa2ise/radios/xmassp.html

If you apply 240V on a 120V bulb, its resistance will go to infinity. AKA burned out. [Linked Image]