2. In the previous section, we studied positive series, but we still lack the tools to analyze series with both positive and negative terms. One of the keys to understanding such series is the concept of absolute convergence. DEFINITION Absolute Convergence

3. Verify that the series converges absolutely. This series converges absolutely because the positive series (with absolute values) is a p-series with p = 2 > 1:absolute values

4. THEOREM 1 Absolute Convergence Implies Convergence The positive series is a p-series with It diverges because p < 1. Therefore, S may converge, but S does not converge absolutely.

5. The series in the previous example does not converge absolutely, but we still do not know whether or not it converges. A series may converge without converging absolutely. In this case, we say that is conditionally convergent. DEFINITION Conditional Convergence An infinite series converges conditionally if converges but diverges. If a series is not absolutely convergent, how can we determine whether it is conditionally convergent? This is often a more difficult question, because we cannot use the Integral Test or the Comparison Test (they apply only to positive series). However, convergence is guaranteed in the particular case of an alternating series.

6. THEOREM 4 Comparison Test Assume that there exists M > 0 such that 0 ≤ a n ≤ b n for n ≤ M.

7. Alternating Series where the terms a n are positive and decrease to zero. An alternating series with decreasing terms. The sum is the signed area, which is at most a 1.

8. THEOREM 2 Leibniz Test for Alternating Series Assume that {a n } is a positive sequence that is decreasing and converges to 0: Then the following alternating series converges: Furthermore, Assumptions Matter The Leibniz Test is not valid if we drop the assumption that a n is decreasing (see Exercise 35).Exercise 35 Example Next Example

9. Show that converges conditionally and that 0 < S < 1. The terms are positive and decreasing, and Therefore, S converges by the Leibniz Test. Furthermore, 0 < S < 1 because a 1 = 1. However, the positive series diverges because it is a p-series with p = ½ < 1. Therefore, S is conditionally convergent but not absolutely convergent. Liebniz Test (A) Partial sums of S = (B) Partial sums

10. The inequality in Theorem 2 gives us important information about the error; it tells us that is less than for all N. THEOREM 3 Let where {a n } is a positive decreasing sequence that converges to 0. Then In other words, the error committed when we approximate S by S N is less than the size of the first omitted term a N+1.

11. Alternating Harmonic Series Show that converges conditionally. Then: (a) Show that (b) Find an N such that S N approximates S with an error less than 10 −3. The terms are positive and decreasing, and Therefore, S converges by the Leibniz Test. The harmonic series diverges, so S converges conditionally but not absolutely. Now, applying Eq. (2), we have For N = 6, we obtain We can make the error less than 10 −3 by choosing N so that Liebniz Test

12. Using a computer algebra system, we find that S 999 ≈ 0.69365. In Exercise 84 of Section 11.7, we will prove that S = ln 2 ≈ 0.69314, and thus we can verify that Exercise 84Section 11.7

13. CONCEPTUAL INSIGHT The convergence of an infinite series depends on two factors: (1) how quickly a n tends to zero, and (2) how much cancellation takes place among the terms. Consider The harmonic series diverges because reciprocals 1/n do not tend to zero quickly enough. By contrast, the reciprocal squares 1/n 2 tend to zero quickly enough for the p-series with p = 2 to converge. The alternating harmonic series converges, but only due to the cancellation among the terms.