If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:2:55

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.7 (EK)

so we have an infinite series here one plus one over two to the fifth plus one over three to the fifth and we just keep on going forever we could write this as the sum from N equals one to infinity of one over N to the fifth power one over N to the fifth power and now you might recognize notice when N is equal to 1 this is 1 over 1 to the 5th that's that over there and we could keep on going now you might immediately recognize this as a p-series and a p-series has the general form of the sum going from N equals 1 to infinity of 1 over N to the P where P is a positive value so in this particular case our P for this P series is equal to 5 P is equal to 5 now you might already recognize under which conditions for AP series does it converge or diverge it's going to converge it's going to converge when your P is greater than 1 which is clearly the case in this scenario right over here our P is clearly greater than 1 we would diverge we would diverge if our P is greater than 0 and less than or equal or less than or equal to 1 this would be a divergence so if this was like 0.9 here or if this was a you know 3/4 then we would be diverging so at least for this one we are convergent let's do another one of these all right so here you might again recognize this as a p-series let me rewrite this infinite sum so this is the sum from N equals 1 to infinity of 1 over C we have a square root of 2 square root of 3 so you could view this as 2 to the 1 half 3 to the 1/2 4 to the 1/2 so it's 1 over N to the 1/2 notice this is when n is equal to 1 1 over the 1/2 is 1 1 over 2 to the 1/2 well that's this right over here and we keep on going on and on and on well in this case we still have a p-series we have 1 over N to some power and that power is positive but notice in this case our P falls between 0 and 1 so 1/2 is our P so P for our P series is equal to 1/2 and that's between 0 and 1 remember we're divergent divergent when our P is greater than 0 and less than or equal to 1 which was clearly the case right over here so this is going to be divergent

AP® is a registered trademark of the College Board, which has not reviewed this resource.