84 CHAPTER 5. INFINITE SERIES OF NUMBERS

In the first case convergence occurs and in the second case, the infinite series diverges. Forthis reason, people will sometimes write ∑

∞k=m ak <∞ to denote the case where convergence

occurs and ∑∞k=m ak = ∞ for the case where divergence occurs. Be very careful you never

think this way in the case where it is not true that all ak ≥ 0. For example, the partialsums of ∑

∞k=1 (−1)k are bounded because they are all either −1 or 0 but the series does not

converge.One of the most important examples of a convergent series is the geometric series.

This series is ∑∞n=0 rn. The study of this series depends on simple high school algebra and

Theorem 4.4.11 on Page 61. Let Sn ≡ ∑nk=0 rk. Then

Sn =n

∑k=0

rk, rSn =n

∑k=0

rk+1 =n+1

∑k=1

rk.

Therefore, subtracting the second equation from the first yields (1− r)Sn = 1−rn+1 and soa formula for Sn is available. In fact, if r ̸= 1,Sn =

1−rn+1

1−r .By Theorem 4.4.11, limn→∞ Sn =1

1−r in the case when |r|< 1. Now if |r| ≥ 1, the limit clearly does not exist because Sn failsto be a Cauchy sequence (Why?). This shows the following.

Theorem 5.1.4 The geometric series, ∑∞n=0 rn converges and equals 1

1−r if |r| < 1and diverges if |r| ≥ 1.

If the series do converge, the following holds.

Theorem 5.1.5 If ∑∞k=m ak and ∑

∞k=m bk both converge and x,y are numbers, then

∞

∑k=m

ak =∞

∑k=m+ j

ak− j (5.1)

∞

∑k=m

xak + ybk = x∞

∑k=m

ak + y∞

∑k=m

bk (5.2)∣∣∣∣∣ ∞

∑k=m

ak

∣∣∣∣∣≤ ∞

∑k=m|ak| (5.3)

where in the last inequality, the last sum equals +∞ if the partial sums are not boundedabove.

Proof: The above theorem is really only a restatement of Theorem 4.4.8 on Page 60and the above definitions of infinite series. Thus

∞

∑k=m

ak = limn→∞

n

∑k=m

ak = limn→∞

n+ j

∑k=m+ j

ak− j =∞

∑k=m+ j

ak− j.

To establish 5.2, use Theorem 4.4.8 on Page 60 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.