When is a sequence convergent or divergent

This will be especially true for sequences that alternate in signs. The following theorem will help with some of these sequences. In this case however the terms just alternate between 1 and -1 and so the limit does not exist. If we do that the sequence becomes,. In this case all we need to do is recall the method that was developed in Calculus I to deal with the limits of rational functions.

We will need to be careful with this one. We will also need to be careful with this sequence. Also, we want to be very careful to not rely too much on intuition with these problems. We will need to use Theorem 2 on this problem. Therefore, since the limit of the sequence terms with absolute value bars on them goes to zero we know by Theorem 2 that,. So, by Theorem 3 this sequence diverges.

We now need to give a warning about misusing Theorem 2. Theorem 2 only works if the limit is zero. If the limit of the absolute value of the sequence terms is not zero then the theorem will not hold. The last part of the previous example is a good example of this and in fact this warning is the whole reason that part is there. Notice that. So, be careful using this Theorem 2. You must always remember that it only works if the limit is zero.

Notes Quick Nav Download. You appear to be on a device with a "narrow" screen width i. Likewise, if the sequence of partial sums is a divergent sequence i. To determine if the series is convergent we first need to get our hands on a formula for the general term in the sequence of partial sums. So, to determine if the series is convergent we will first need to see if the sequence of partial sums,. The limit of the sequence terms is,. So, as we saw in this example we had to know a fairly obscure formula in order to determine the convergence of this series.

In general finding a formula for the general term in the sequence of partial sums is a very difficult process. We will continue with a few more examples however, since this is technically how we determine convergence and the value of a series. This is actually one of the few series in which we are able to determine a formula for the general term in the sequence of partial fractions. Therefore, the series also diverges. Again, do not worry about knowing this formula.

The sequence of partial sums is convergent and so the series will also be convergent. The value of the series is,. As we already noted, do not get excited about determining the general formula for the sequence of partial sums.

Two of the series converged and two diverged. Notice that for the two series that converged the series term itself was zero in the limit. This will always be true for convergent series and leads to the following theorem.

Then the partial sums are,. Be careful to not misuse this theorem! This theorem gives us a requirement for convergence but not a guarantee of convergence. In other words, the converse is NOT true. So this is y is equal to a sub n. And we keep going on and on and on.

So you see the points, they kind of jump around, but they seem to be getting closer and closer and closer to 0. Which would make us ask a very natural question-- what happens to a sub n as n approaches infinity? Or another way of saying that is, what is the limit-- let me do this in a new color-- of a sub n as n approaches infinity?

Well, let's think about if we can define a sub n explicitly. So we can define this sequence as a sub n where n starts at 1 and goes to infinity with a sub n equaling-- what does it equal? Well, if we ignore sign for a second, it looks like it's just 1 over n. But then we seem like we oscillate in signs.

We start with a positive, then a negative, positive, negative. So we could multiply this times negative 1 to the-- let's see. If we multiply it times negative 1 to the n, then this one would be negative and this would be positive. But we don't want it that way. We want the first term to be positive. So we say negative 1 to the n plus 1 power. And you can verify this works. When n is equal to 1, you have 1 times negative 1 squared, which is just 1, and it'll work for all the rest.

So we could write this as equaling negative 1 to the n plus 1 power over n. And so asking what the limit of a sub n as n approaches infinity is equivalent to asking what is the limit of negative 1 to the n plus 1 power over n as n approaches infinity is going to be equal to? Remember, a sub n, this is just a function of n.