154 CHAPTER 6. INFINITE SERIES

Proof: It follows {∑nk=m ak}∞

n=m is an increasing sequence since ∑n+1k=m ak −∑

nk=m ak =

an+1 ≥ 0. If the sequence of partial sums is bounded above, then this sequence of partialsums must converge to S ≡ sup{∑

nk=m ak : n ≥ m} by Lemma 6.1.3. If the sequence of

partial sums is not bounded, then it cannot converge because if it converged to S, then forall n large enough, |∑n

k=m ak −S|< 1, and for all such n,∑nk=m ak ∈ (1−S,1+S) , and there

are only finitely many other terms so {∑nk=m ak} would need to be bounded.

In the case where ak ≥ 0, the above proposition shows there are only two alternativesavailable. Either the sequence of partial sums is bounded above or it is not bounded above.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 3.3.10 on Page 84. Let Sn ≡ ∑nk=0 rk. Then Sn = ∑

nk=0 rk, rSn = ∑

nk=0 rk+1 =

∑n+1k=1 rk.Therefore, subtracting the second equation from the first yields (1− r)Sn = 1−

rn+1and so a formula for Sn is available. In fact, if r ̸= 1,Sn =1−rn+1

1−r . By Theorem 3.3.10,limn→∞ Sn =

11−r in the case when |r| < 1. Now if |r| ≥ 1, the limit clearly does not exist

because Sn fails to be a Cauchy sequence (Why?) so by Theorem3.7.3 it cannot converge.This shows the following.

Theorem 6.1.5 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 6.1.6 If ∑∞k=m ak and ∑

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

∞

∑k=m

ak =∞

∑k=m+ j

ak− j (6.1)

∞

∑k=m

xak + ybk = x∞

∑k=m

ak + y∞

∑k=m

bk (6.2)

∣∣∣∣∣ ∞

∑k=m

ak

∣∣∣∣∣≤ ∞

∑k=m

|ak| (6.3)

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 3.3.7 on Page 83and 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.