# AP Calculus: Series Divergence and Convergence

AP Calculus Prep: An infinite series is said to converge if, as the index of the series approaches infinity, the series approaches a limit, L. As you may recall, the limit L has to be a real number – infinity is not among the allowed values. Series that do not converge are said to diverge. To put it mathematically,

For the series to be convergent,  the sum as n approaches infinity. If no such number L exists, then the series is divergent.

Let us now think about the possibilities for the sequence of the terms :could increase, decrease, or stay the same as i increases. If the terms are constant, i.e., independent of i, then the left hand side above would be equal to that constant times infinity, which is infinity. Since L has to be finite, the limit doesn’t exist and we can say that the series diverges.

If the terms increase, we can find a constant value c which is always less than . We therefore know that the sum of all n  is greater than . Since we just proved that an infinite series of constant terms tends to infinity (i.e., diverges), so must a series whose terms are all greater than the constant.

By the way, this type of reasoning is called a comparison test. We compared the terms of the series we were testing (the one with increasing terms) to a series that we knew diverged and had every term smaller than the terms of the series being tested. Since our test series is greater than the other series, our test series must also diverge.

By contrast, if we wanted to prove that a series was convergent, we would want to compare it to a known convergent series whose terms were all greater than those of the series we are testing.

So if we see the terms increasing or staying constant, we know right away that the series diverges. Now what about if the terms are decreasing? Is that sufficient to guarantee convergence of the series?

Let’s consider the series , also known as the harmonic series. We can write out the terms of the harmonic series as follows:

1 + ½ + 1/3 + ¼ + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + …..

We can now invent another series with terms always smaller than those of the harmonic series:

1 + ½ + ¼ + ¼ + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + …..

The second series can also be written as

1 + ½ + (1/4 + ¼) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + ….) ….

Adding up the terms in the brackets, we get

1 + ½ + ½ + ½ + ….

So we see that the second series is a series of constant terms, which we know diverges. Therefore the harmonic series must also diverge.  Getting back to the previous question, we can now say that the fact that the terms of a series are decreasing  is not sufficient to guarantee its convergence.

Good luck on the AP Calculus Exam!