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.

# Alternating series test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.10 (EK)

## Video transcript

let's now expose ourselves to another test of convergence and that's the alternating series test and I'll explain the alternating series test and I'll apply it to an actual series while I do it to make the the explanation of the alternating series test a little bit more concrete so let's say that I have some series some infinite series let's say it goes from N equals K to infinity of a sub N and let's say I can write it as or I can rewrite a sub n so let's say a sub n I can write so a sub n is equal to negative 1 to the end times B sub N or a sub n is equal to negative 1 to the n plus 1 times B sub n where where B sub n is greater than or equal to 0 for all the ends we care about so for all of these integer ends greater than or equal to K so if all of these things if all of these things are true and we know two more things and we know number one the limit as n approaches infinity of B sub n is equal to 0 and number 2 B sub n is a decreasing sequence decreasing decreasing sequence then that lets us know that the original infinite series the original infinite series is going to converge so this might seem a little bit abstract right now let's actually you show let's let's use this with an actual series to make it a little bit more a little bit more concrete so let's say that I had the series let's say I had the series from N equals 1 to infinity of negative 1 to the N over N and we could write it out just to make this series a little bit more concrete when n is equal to 1 this is going to be negative 1 to the 1 power actually let's just make this a little bit let's make this a little more interesting let's make this negative 1 to the n plus 1 so what n is equal to 1 this is going to be negative 1 squared over 1 which is going to be 1 and then when n is 2 it's gonna be negative 1 to the third power which is going to be negative 1/2 so it's minus 1/2 plus 1/3 minus 1/4 plus minus and it just keeps going on and on and on forever now can we rewrite this a sub n like this well sure the negative 1 to the n plus 1 is actually explicitly called out we can rewrite we can rewrite our a sub n so let me do that so negative so a sub n which is equal to negative 1 to the n plus 1 over n this is clearly the same thing as negative 1 to the n plus 1 times 1 over n which is which we can then say this thing right over here could be our B sub n so this right over here is our B sub N and we can verify that our B sub n is going to be greater than or equal to 0 for all the ends we care about so our B sub n is equal to 1 over n now clearly this is going to be greater than or equal to 0 for any for any positive n now what's the limit as B sub n what's the limit of B sub n as n approaches infinity the limit let me just write 1 over N 1 over n as n approaches infinity is going to be equal to 0 so we satisfy the first constraint and then this is clearly a decreasing sequences and as n increases the denominators are going to increase and it was a larger denominator you're going to have a lower value so we can also say 1 over N is a decreasing decreasing sequence for the ends that we care about so this is this satisfies this is satisfied as well and so based on that this thing this thing is always this thing right over here is always greater equal to zero the limit as one over n R's are B sub n as n approaches infinity is going to be 0 it's a decreasing sequence therefore we can say that our original series actually converges so N equals 1 to infinity of negative 1 to the n plus 1 over N and that's kind of interesting because we've already seen that if all of these were positive if all of these terms are positive we just have the harmonic series and that one didn't converge but this one did it putting these negatives here do the trick and actually we can prove this one over here converges using other techniques and maybe if we have time actually in particular the limit comparison test I'll just throw that out there in case you are curious so this is a pretty powerful tool it looks a little bit but like that divergence test but remember the divergence test is really is only useful if you want to show something diverges if the limit of if the limit of your terms do not approach zero then you say okay that thing is going to diverge this thing is useful because you can actually prove convergence now once again if something does not pass the alternating series test that does not necessarily mean that it diverges it just means that you couldn't use the alternating series test to prove that it converges
AP® is a registered trademark of the College Board, which has not reviewed this resource.