8.3 Series with Positive and Negative Terms 481Theorem 8.3.10 Given any series I: an of real numbers,
(a) I: an converges absolutely~ both I: a;t and I: a;;: converge.
(b) If I: an converges absolutely then it converges.
( c) If I: an converges conditionally then both I: a;t and I: a;;: diverge to +oo.
( d) If one of the series I: a;t, I: a;;: converges and the other diverges, then
I: an diverges.Proof. First note that because of Lemma 8.3.9,(*) Lan= _L(a;t - a;;:) and (**) L lanl = _L(a;t +a;;:).
(a) If I: lanl converges, then from the inequalities 0 :::; a;t :::; lanl and
0 :::; a;;: :::; lanJ, together with the comparison test for nonnegative series, both
I: a;t and I: a;;: converge.
Conversely, if both I: a;t and I: a;;: converge, then (**) together with The-
orem 8.1.12 assures us that I: Jani converges.
(b) If I: Ian I converges, then by Part (a) both I: a;t and I: a;;: converge,
so (*) together with Theorem 8.1.12 assures us that _Lan = _L(a;t - a;;:)
converges.
( c) Suppose I: an converges, but not absolutely. For contradiction, suppose
that one of the series I: a;t, I: a;;: converges. Without loss of generality, suppose
the second of these two converges. That is,
_L(a;t - a;;:) and I: a;;: converge.
Then, by Theorem 8.1.12, _L[(a;t - a;;:) +a;;:] converges; i.e., I: a;t con-
verges. But then, by (a), _Lan converges absolutely. Contradiction.
(d) Exercise 3. •Theorem 8.3.11 (Generalized Triangle Inequality) If I: an converges
absolutely, then J L an I :S L I an I·
Proof. Suppose I: an converges absolutely. Then, by Theorem 8.3. 10 (a),
both I: a;t and I: a;;: converge, and using Lemma 8.3.9,
:::; II: a;t I + II: a;;: I (by the ordinary triangle inequality)
= I: a;t + I: a;;: = _L( a;t + a;;:) = I: Jan J. •