Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

900 ratings

The Ohio State University

900 ratings

Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Convergence Tests

In this third module, we study various convergence tests to determine whether or not a series converges: in particular, we will consider the ratio test, the root test, and the integral test.

- Jim Fowler, PhDProfessor

Mathematics

The ratio test is looking like a great test. Does it always work? [SOUND]. Let's recall what the ratio test says. I'm going to be considering a series the sum of a sub n and goes from zero to infinity. And I'm going to assume that all of the terms that I'm adding up, all of the a sub ns, are non-negative. So a sub n is greater than or equal to 0. And the ratio test tells me to consider the limit of the ratio of subsequent terms, so I'm taking the limit, as n approaches infinity of a sub n plus 1 over a sub n, and I'm calling that limit L.

And then what does the ratio test tell me? Well the ratio test tells me that if that limit is less than 1, then the series converges. If that limit is bigger than 1, then the series diverges. But if the limit is equal to 1 then the ratio test is inconclusive. So what happens when l equals one? Well supposedly the ratio test doesn't say anything in that case. But if we listen really carefully, is there something that we can get out of the ratio test even when l equals one? No. Whoa, okay, but why not? Well there are cases where l equals 1 and the series converges. But then there are also cases where L equals one and the series diverges. Can we actually see some of these examples? Here's an example but maybe a silly example where L is equal to 1 but the series will end up diverging.

So let's try this. Here's the example, it's the sum, n goes from 0 to infinity of the number 1. [LAUGH] well, I mean this, this definitely divergences, right? If you keep adding up 1 plus 1 plus 1, right, that's not a finite number. So this is a divergent series. But of course, in this case, what's a sub n? Well a sub n doesn't depend on n, it's just always 1. And so the limit of the n + 1th term over the nth term as n goes to infinity is just 1, because a sub n plus 1 and a sub n are both 1. What about an example where the series converges even though L equals 1? So on an example where the L in the ratio test is 1, but the series ends up converging even though the ratio test doesn't tell us that. So let's see an example. An example of this is the sum and goes from 1 to infinity of 1 over n squared. THis series we've already seen to converge. It converges by, let's say, the p series test. It's a p series where p equals 2. But, what is l in this case. Well, the nth term in the series a sub n, is 1 over n squared. So l is the limit of a sub n plus1 over a sub n, as n goes to infinity. And in that case, that's the limit as n goes to infinity of 1 over n plus 1 squared. Divided by 1 over n squared and that limit is indeed 1. But how do I know that the limit is 1? Well, to compute this limit, I could first rewrite this as the limit as n approaches infinity 1 over what, I must gotta move the n plus 1 squared in the denominator so it will be n plus 1 squared over n squared now [INAUDIBLE] expand it out to the n plus 1 squared this be the limit and as n approaches infinity 1 over n squared plus 2n plus 1 that's what I got expanded n plus 1 squared, divided by n squared, And this is the sum here in the numerator and I got split this up into three separate fractions. This is the limit as n goes to infinity 1 over n squared over n squared, plus two n over n squared plus one over n squared. Now I could simplify this a bit more too. This is the limit n goes to infinity of one over n squared over n squared is one. 2 n over n squared is 2 over and and 1 over n squared I'll just write as one over n squared. Well what's the limit? When and is very large, this is very close to 0 and this is very close to 0. So when n is very large, this is 1 over 1 plus a number close to 0 plus a number close to 0. This limit Is 1, which is exactly what I'm claiming. So when we say that the ratio test is silent when l equals 1, it's not for a lack of understanding. There really are series that converge and series that diverge. In both cases, l equals 1. [NOISE]

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