6.2. ABSOLUTE CONVERGENCE 155

To establish 6.2, use Theorem 3.3.7 on Page 83 to write

∞

∑k=m

xak + ybk = limn→∞

n

∑k=m

xak + ybk = limn→∞

(x

n

∑k=m

ak + yn

∑k=m

bk

)

= x∞

∑k=m

ak + y∞

∑k=m

bk.

Formula 6.3 follows from the observation that, from the triangle inequality,∣∣∣∣∣ n

∑k=m

ak

∣∣∣∣∣≤ ∞

∑k=m

|ak|

and so |∑∞k=m ak|= limn→∞ |∑n

k=m ak| ≤ ∑∞k=m |ak| .

Recall that if limn→∞ An = A, then limn→∞ |An|= |A|.

Example 6.1.7 Find ∑∞n=0( 5

2n +63n

).

From the above theorem and Theorem 6.1.5, ∑∞n=0( 5

2n +63n

)=

5∞

∑n=0

12n +6

∞

∑n=0

13n = 5

11− (1/2)

+61

1− (1/3)= 19.

The following criterion is useful in checking convergence. All it is saying is that theseries converges if and only if the sequence of partial sums is Cauchy. This is what thegiven criterion says. It is just a re-statement of Theorem 3.7.3 on Page 93. It is not newinformation.

Theorem 6.1.8 Let {ak} be a sequence of points in R. The sum ∑∞k=m ak converges

if and only if for all ε > 0, there exists nε such that if q ≥ p ≥ nε , then∣∣∣∣∣ q

∑k=p

ak

∣∣∣∣∣< ε. (6.4)

Proof: Suppose first that the series converges. Then {∑nk=m ak}∞

n=m is a Cauchy se-quence by Theorem 3.7.3 on Page 93. Therefore, there exists nε > m such that if q ≥p−1 ≥ nε > m, ∣∣∣∣∣ q

∑k=m

ak −p−1

∑k=m

ak

∣∣∣∣∣=∣∣∣∣∣ q

∑k=p

ak

∣∣∣∣∣< ε. (6.5)

Next suppose 6.4 holds. Then from 6.5 it follows upon letting p be replaced with p+1that {∑

nk=m ak}∞

n=m is a Cauchy sequence and so, by Theorem 3.7.3, it converges. By thedefinition of infinite series, this shows the infinite sum converges as claimed.

6.2 Absolute ConvergenceAbsolute convergence is the best kind. It says that if you replace each term with its absolutevalue, the resulting series converges.