Alternating Series

The convergence tests described in the last section apply only to series with positive terms. Of particular importance are alternating series, whose terms alternate in sign. For example:

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots = \sum_{n = 1}^{\infty} (- 1)^{n - 1} \frac{1}{n}

We see from these examples that the nth term of an alternating series is of the form

a_n = (−1)ⁿ⁻¹ b_n

where b_n is a positive sequence (we have b_n = a_n). The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges.

Proposition 3.4.1 (Leibniz Criterion). An alternating series Σ_n (−1)^{n − 1} b_n is convergent if it satisfies the following conditions:

b_n+1 ≤ b_n, ∀n (the sequence b_n is decreasing)
lim_{n ⟶ ∞} b_n = 0.

Proof. To prove this, we examine the even partial sums

s_2n = b₁ − b₂ + b₃ − ... − b_2n
s_{2n + 2} = s_2n + (b_{2n + 1} − b_{2n + 2}).

Since b_{2n +1} > b_{2n + 2}, we have

s_{2n + 2} > s_2n

On the other hand,

s_{2n + 2} = a₁ − (b₂ − b₃) − (b₄ − b₅) − ... − b_{2n + 2}

Eeach pair in the brackets is positive, then

s_{2n + 2} < b₁

With the even partial sums bounded s_2n < s_{2n + 2} < b₁; therefore the sequence {s_2n} of even partial sums is increasing and bounded above. It is therefore convergent by the Monotone Sequence Theorem. Let’s call its limit s, that is,

lim_{n ⟶ ∞} s_2n = s

Now we compute the limit of the odd partial sums:

lim_{n ⟶ ∞} s_2n+1 = lim_{n ⟶ ∞} (s_2n + b_2n+1) = lim_{n ⟶ ∞} s_2n + lim_{n ⟶ ∞} b_2n+1
= s + 0 = s

Since both the even and odd partial sums converge to s, we have lim_{n ⟶ ∞} s_n = s. □

Definition 3.4.2. A series Σ a_n is said to be absolutely convergent if the series of the moduli of its terms Σ |a_n| is convergent. □

Theorem 3.4.3. If a series is absolutely convergent, then it is convergent.

Proof. Since Σ |a_n| is convergent, it satisfies Cauchy criterion, that is, for all ε > 0 there is some N such that

\sum_{k = m + 1}^{n} | a_{k} | < ε, n > m \geq N

Since

| \sum_{k = m + 1}^{n} a_{k} | \leq \sum_{k = m + 1}^{n} | a_{k} | < ε, n > m \geq N

Then the series Σ a_n satisfies as well the Cauchy criterion, therefore it converges. □

In other words, absolute convergence implies convergence, but not vice versa as the following example illustrates.

Example 3.4.3. The series Σ (−1)ⁿ/n^α, with α > 1 is absolutely convergent. Indeed we have

|(−1)ⁿ/n^α| = 1/n^α

The series Σ 1/n^α is converge for α > 1. If α = 1, nothing we can conclude on its convergence, since the series Σ 1/n (harmonic series) is divergent. ■

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