Earlier in Definition 3.2.1 on Page 213 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 this
section.
Definition 10.1.1Define
∞∑ ∑n
ak ≡ nl→im∞ ak
k=m k=m
whenever the limit exists and is finite. In this case the series is said to converge. If itdoes not converge, it is said to diverge. The sequence
∑
{ nk=m ak}
_{n=m}^{∞}inthe above is called the sequence of partial sums.This is always the definition.Here it is understood that the a_{k}are in ℝ, but it is the same definition in anysituation.
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 partial
sums. 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.
= r. In the first case, suppose r < ∞. Then letting
ε > 0 be given, there exists n such that A_{n}∈ (r − ε,r]. Since
{An}
is increasing, it
follows if m > n, then r − ε < A_{n}≤ A_{m}≤ r and so lim_{n→∞}A_{n} = r as claimed. In the
case where r = ∞, then if a is a real number, there exists n such that A_{n}> a. Since
{Ak}
is increasing, it follows that if m > n, A_{m}> a. But this is what is meant by
lim_{n→∞}A_{n} = ∞. The other case is that r = −∞. But in this case, A_{n} = −∞
for all n and so lim_{n→∞}A_{n} = −∞. The other claim is shown the same way.
■
Proposition 10.1.4Let a_{k}≥ 0. Then
∑n
{ k=m ak}
_{n=m}^{∞}is an increasingsequence. If this sequence is bounded above, then∑_{k=m}^{∞}a_{k}converges and its valueequals
{ }
∑n
sup ak : n = m, m + 1,⋅⋅⋅ .
k=m
When the sequence is not bounded above,∑_{k=m}^{∞}a_{k}diverges.However, in this case,people sometimes write∑_{k=m}^{∞}a_{k} = ∞.
Proof: It follows
∑n
{ k=m ak}
_{n=m}^{∞} is an increasing sequence because
n∑+1 ∑n
ak − ak = an+1 ≥ 0.
k=m k=m
If the sequence of partial sums is bounded above, then this sequence of partial sums must
converge to S ≡ sup
∑n
{ k=m ak : n ≥ m}
by Lemma 10.1.3. If the sequence of partial
sums is not bounded, then it cannot converge because if it converged to S, then for all n
large enough,
∑n
| k=m ak − S|
< 1, and for all such n,∑_{k=m}^{n}a_{k}∈
(1− S,1 + S)
, and
there are only finitely many other terms so
∑n
{ k=m ak}
would need to be bounded.
■
In the case where a_{k}≥ 0, the above proposition shows there are only two
alternatives available. 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. For this reason, people will sometimes write
∑_{k=m}^{∞}a_{k}< ∞ to denote the case where convergence occurs and ∑_{k=m}^{∞}a_{k} = ∞ for
the case where divergence occurs. Be very careful you never think this way in
the case where it is not true that all a_{k}≥ 0. For example, the partial sums 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}^{∞}r^{n}. The study of this series depends on simple high school algebra
and Theorem 3.2.9 on Page 225. Let S_{n}≡∑_{k=0}^{n}r^{k}. Then
∑n k ∑n k+1 n∑+1 k
Sn = r ,rSn = r = r .
k=0 k=0 k=1
Therefore, subtracting the second equation from the first yields
(1− r)Sn = 1− rn+1
and so a formula for S_{n} is available. In fact, if r≠1,
≥ 1, the limit
clearly does not exist because S_{n} fails to be a Cauchy sequence (Why?) so by Corollary
6.1.6 it cannot converge. This shows the following.
The following criterion is useful in checking convergence. All it is saying is that the
series converges if and only if the sequence of partial sums is Cauchy. This is what the
given criterion says. It is just a re-statement of Corollary 6.1.6 on Page 422. It is not new
information.
Theorem 10.1.8Let
{ak}
be a sequence of points in ℝ. The sum∑_{k=m}^{∞}a_{k}convergesif and only if for all ε > 0, there exists n_{ε}such that if q ≥ p ≥ n_{ε},then
| |
||∑q ||
|| ak||< ε. (10.4)
|k=p |
(10.4)
Proof: Suppose first that the series converges. Then
∑n
{ k=m ak}
_{n=m}^{∞} is a Cauchy
sequence by Corollary 6.1.6 on Page 422. Therefore, there exists n_{ε}> m such that if
q ≥ p − 1 ≥ n_{ε}> m,
Next suppose 10.4 holds. Then from 10.5 it follows upon letting p be replaced with
p + 1 that
{∑nk=m ak}
_{n=m}^{∞} is a Cauchy sequence and so, by Corollary 6.1.6, it
converges. By the definition of infinite series, this shows the infinite sum converges as
claimed. ■