Chapter 6

Infinite Series6.1 Basic Considerations

Earlier in Definition 3.3.1 on Page 82 the notion of limit of a sequence was discussed.There is a very closely related concept called an infinite series which is dealt with in thissection.

Definition 6.1.1 Define ∑∞k=m ak ≡ limn→∞ ∑

nk=m ak whenever the limit exists and is

finite. In this case the series is said to converge. If it does not converge, it is said to diverge.The sequence {∑

nk=m ak}∞

n=m in the above is called the sequence of partial sums. This isalways the definition. Here it is understood that the ak are in R, but it is the same definitionin any situation.

From this definition, it should be clear that infinite sums do not always make sense.Sometimes they do and sometimes they don’t, depending on the behavior of the partialsums. As an example, consider ∑

∞k=1 (−1)k. The partial sums corresponding to this symbol

alternate between −1 and 0. Therefore, there is no limit for the sequence of partial sums.It follows the symbol just written is meaningless and the infinite sum diverges.

Example 6.1.2 Find the infinite sum, ∑∞n=1

1n(n+1) .

Note 1n(n+1) =

1n −

1n+1 and so ∑

Nn=1

1n(n+1) = ∑

Nn=1( 1

n −1

n+1

)=− 1

N+1 +1. Therefore,

limN→∞

N

∑n=1

1n(n+1)

= limN→∞

(− 1

N +1+1)= 1.

Lemma 6.1.3 If {An} is an increasing sequence in [−∞,∞], then sup{An}= limn→∞ AnIf {An} is a decreasing sequence, then inf{An}= limn→∞ An.

Proof: Let sup({An : n ∈ N}) = r. In the first case, suppose r < ∞. Then letting ε > 0be given, there exists n such that An ∈ (r − ε,r]. Since {An} is increasing, it follows ifm > n, then r − ε < An ≤ Am ≤ r and so limn→∞ An = r as claimed. In the case wherer = ∞, then if a is a real number, there exists n such that An > a. Since {Ak} is increasing,it follows that if m > n, Am > a. But this is what is meant by limn→∞ An = ∞. The othercase is that r =−∞. But in this case, An =−∞ for all n and so limn→∞ An =−∞. The otherclaim is shown the same way.

Proposition 6.1.4 Let ak ≥ 0. Then {∑nk=m ak}∞

n=m is an increasing sequence. If thissequence is bounded above, then ∑

∞k=m ak converges and its value equals

sup

{n

∑k=m

ak : n = m,m+1, · · ·

}.

When the sequence is not bounded above, ∑∞k=m ak diverges. However, in this case, people

sometimes write ∑∞k=m ak = ∞.

153