 One often useful feature of mathematics is that existing work can be used in other contexts by introducing a new concept. In this case, we'll introduce the concept of conditional convergence. Suppose I have a series of terms not always positive and not necessarily alternating. However, the term itself will be some place between the absolute value of the term and minus the absolute value of the term. And that means we can trap the partial sums between two series, the sum of the absolute values and the sum of the negative absolute values. And this means that if this sum of absolute values is finite, then so is this sum of negative absolute values. And that means the series itself must also converge. And this leads to the following theorem. If our series of absolute values converges, then our series of terms also converges. Now it's possible for the reverse not to hold true. In particular, it's possible for the series to converge, but for the series of absolute values to diverge. And this leads to the notion of conditional convergence. Suppose I have a series that converges, but the series of absolute values diverge. We say that our series is conditionally convergent and that it converges conditionally. On the other hand, if our series of absolute values converges, we say that the original series is absolutely convergent and it converges absolutely. And here's a useful thing to remember. By our theorem, if a series converges absolutely, it's guaranteed to converge conditionally. On the other hand, if our series does not converge absolutely, we must check for conditional convergence. So let's consider this series from 1 to infinity of minus 1 to power n plus 1 over n. Now it's useful to remember that our ratio test, our root test, and our integral test actually do test for absolute convergence. So a general strategy is to check for absolute convergence first using one of our standard tests, and only if it fails do we worry about whether the series might be conditionally convergent. Since the absolute value of the terms of the series are given by the function f of x equals 1 over x, we could use the integral test. So we'll integrate 1 over x from 1 to infinity, and we find that this integral diverges, and so this means the series does not converge absolutely. And we might summarize our result. By the integral test, our series of 1 over n does not converge, so the series is not absolutely convergent. It is possible that the series as written will converge, so we'll have to check that out. Since our series is an eventually alternating, eventually decreasing series, we can apply the alternating series test. And in this case we note that limit as n goes to infinity of 1 over n is 0, so the series converges. And so by the alternating series test, our original series converges, so our series is conditionally convergent. How about the convergence of the series sin n over n squared minus 2? So the first thing to recognize is this is not an alternating series, so we can't use the alternating series test. But maybe we won't have to. Remember, it's useful to begin by checking for absolute convergence. The first important observation to make is the absolute value of sin of n will be less than or equal to 1. So we could test for absolute convergence using the comparison or the limit comparison test. Actually, we'll use both. First, our series of positive terms is going to be less than or equal to the absolute value of 1 over n squared minus 2. Unfortunately, we don't know what happens to the series with terms 1 over n squared minus 2. But we do know that the series 1 over n squared converges, so let's use the limit comparison test to compare what happens to the series 1 over n squared minus 2 and the series 1 over n squared. So we'll find the limit as n goes to infinity of the ratio of the terms of the two series, which turns out to be 1, and so we know that both series converge. So we might summarize our findings as follows. We know that the absolute value of the terms of our series are less than or equal to the absolute value of 1 over n squared minus 2. By the limit comparison test, we know that the series with terms absolute value of 1 over n squared minus 2 converges. So by the comparison test, the series with terms absolute value of sin n over n squared minus 2 converges. And since the series of absolute value converges, then our series of not absolute values will converge absolutely.