Attaching a LED to PIC's Rx pin
(At least some) PICs can't drive much current(*), but also for the RxD pin you better use a transistor to drive the LED, since you'll avoid loading the transmitter at the other end (probably a MAX3232 or similar?).



Connect the input "Q" to the TxD/RxD line. A typical general purpose transistor will have a gain of about 100, then 1 mA base current is enough to get 20 mA collector current. 

For a 5 V bus and power supply:
choose \$R_B\$ = 3.9 k&Omega;, then the base current will be (5 V - 0.7 V)/ 3.9 k&Omega; = 1.1 mA. To limit the collector current to 20 mA (typical indicator LED) \$R\$ should be (5 V - 2 V)/ 20 mA = 150 &Omega;.

For a 3.3 V bus and power supply, use the same equations, replacing 5 V by 3.3 V, then your resistor values will be 2.2 k&Omega; and 47 &Omega; resp.


A MOSFET like AndrejaKo suggests is a good alternative, but make sure you have a logic level gate type, with a maximum gate threshold voltage somewhat below the bus voltage. (There are logic level gate FETs where that can be as high as 4 V and then you won't get enough drain current with a 3.3 V bus voltage.) The real advantage of the FET is that it needs hardly any drive current, but since we only need a mA for the BJT we won't have any problems with that either.



(*) This random PIC controller specifies a 700 mV drop at only 3 mA output current, that's a 230 &Omega; output resistance. A 2 V LED directly driven from a 3.3 V output will drop the output by 1 V at only 4 mA. Most indicator LEDs are specified for 20 mA.