How does a capacitor reduce voltage drop in a DC motor start?
Your voltage source is assumed to have some resistance. The capacitor would provide some of the current during the start-up surge. 







simulate this circuit &ndash; Schematic created using CircuitLab

So, say it takes 0.5 second for the motor to wind up and it draws 0.5A during the surge and 100mA after (of course it will vary more smoothly than that). Then you would like the voltage to drop less than 1V during the surge. Let's assume all the current is supplied by the capacitor. 

So \$C = \frac{I_{start}\cdot T_{start}}{\Delta V}\$ = 0.25F (250,000uF). 

As you can see, you'll need a pretty large capacitor for even a small motor. 



Putting it in series would just cause the motor to jog for a bit and then stop when the switch is closed, assuming the capacitor was discharged to begin with. It could only reduce the current available to the motor (again assuming it was discharged). Once the cap is charged there is no more current and the motor stops.