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

The series converges absolutely if converges.n na a

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:

THEOREM 1 Absolute Convergence Implies Convergence

converges also converges.n na a

1

1

1 1 1 1Does + converge absolutely?1 2 3

n

n

Sn

The positive series is a p-series with 1 .2

p

It diverges because p < 1. Therefore, S may converge, but S does not converge absolutely.

1na

n

The series in the previous example does not converge absolutely, but we still do not know whether or not it converges. A series namay converge without converging absolutely. In this case, we say that

is conditionally convergent.naDEFINITION Conditional Convergence An infinite series naconverges conditionally if converges but diverges.na na

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.

THEOREM 4 Comparison Test Assume that there exists M > 0 such that 0 ≤ an ≤ bn for n ≤ M.

1 1

1 1

i converges converges.

ii diverges diverges.

n nn n

n nn n

b a

a b

Alternating Series

where the terms an are positive and decrease to zero.

An alternating series with decreasing terms. The sum is the signed area, which is at most a1.

1 1S S a

THEOREM 2 Leibniz Test for Alternating Series Assume that {an} 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 an is decreasing (see Exercise 35).

4 5. . Let 2i e N S S S

ExampleNext Example

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 a1 = 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.

1

1

1 nn

n

S a

Liebniz Test(A) Partial sums of S = 1

1

1 nn

n

a

(B) Partial sums 1

1n n

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 {an} is a positive decreasing sequence that converges to 0. Then

In other words, the error committed when we approximate S by SN is less than the size of the first omitted term aN+1.

Alternating Harmonic Series Show that

converges conditionally. Then:

(a) Show that

(b) Find an N such that SN 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

Using a computer algebra system, we find that S999 ≈ 0.69365. In Exercise 84 of Section 11.7, we will prove that S = ln 2 ≈ 0.69314, and thus we can verify that

CONCEPTUAL INSIGHT The convergence of an infinite series nadepends on two factors: (1) how quickly an 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/n2 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.