3Convergence of infinite sums
IA Analysis I
3.2 Absolute convergence
Here we’ll consider two stronger conditions for convergence — absolute conver-
gence and unconditional convergence. We’ll prove that these two conditions are
in fact equivalent.
Definition (Absolute convergence). A series
P
a
n
converges absolutely if the
series
P
|a
n
| converges.
Example. The series
P
(−1)
n+1
n
= 1
−
1
2
+
1
3
−
1
4
+
···
converges, but not
absolutely.
To see the convergence, note that
a
2n−1
+ a
2n
=
1
2n − 1
−
1
2n
=
1
2n(2n − 1)
.
It is easy to compare with 1
/n
2
to get that the partial sums
S
2n
converges. But
S
2n+1
− S
2n
= 1/(2n + 1) → 0, so the S
2n+1
converges to the same limit.
It does not converge absolutely, because the sum of the absolute values is
the harmonic series.
Lemma. Let
P
a
n
converge absolutely. Then
P
a
n
converges.
Proof.
We know that
P
|a
n
|
converges. Let
S
N
=
P
N
n=1
a
n
and
T
N
=
P
N
n=1
|a
n
|.
We know two ways to show random sequences converge, without knowing
what they converge to, namely monotone-sequences and Cauchy sequences. Since
S
N
is not monotone, we shall try Cauchy sequences.
If p > q, then
|S
p
− S
q
| =
p
X
n=q+1
a
n
≤
p
X
n=q+1
|a
n
| = T
p
− T
q
.
But the sequence
T
p
converges. So
∀ε >
0, we can find
N
such that for all
p > q ≥ N, we have T
p
− T
q
< ε, which implies |S
p
− S
q
| < ε.
Definition (Unconditional convergence). A series
P
a
n
converges uncondition-
ally if the series
P
∞
n=1
a
π(n)
converges for every bijection
π
:
N → N
, i.e. no
matter how we re-order the elements of a
n
, the sum still converges.
Theorem. If
P
a
n
converges absolutely, then it converges unconditionally.
Proof. Let S
n
=
P
N
n=1
a
π(n)
. Then if p > q,
|S
p
− S
q
| =
p
X
n=q+1
a
π(n)
≤
∞
X
n=q+1
|a
π(n)
|.
Let ε > 0. Since
P
|a
n
| converges, pick M such that
P
∞
n=M+1
|a
n
| < ε.
Pick N large enough that {1, ··· , M} ⊆ {π(1), ··· , π(N)}.
Then if n > N, we have π(n) > M. Therefore if p > q ≥ N, then
|S
p
− S
q
| ≤
p
X
n=q+1
|a
π(n)
| ≤
∞
X
n=M+1
|a
n
| < ε.
Therefore the sequence of partial sums is Cauchy.
The converse is also true.
Theorem. If
P
a
n
converges unconditionally, then it converges absolutely.
Proof.
We will prove the contrapositive: if it doesn’t converge absolutely, it
doesn’t converge unconditionally.
Suppose that
P
|a
n
|
=
∞
. Let (
b
n
) be the subsequence of non-negative
terms of
a
n
, and (
c
n
) be the subsequence of negative terms. Then
P
b
n
and
P
c
n
cannot both converge, or else
P
|a
n
| converges.
wlog,
P
b
n
=
∞
. Now construct a sequence 0 =
n
0
< n
1
< n
2
< ···
such
that ∀k,
b
n
k−1
+1
+ b
n
k−1
+2
+ ··· + b
n
k
+ c
k
≥ 1,
This is possible because the
b
n
are unbounded and we can get it as large as we
want.
Let π be the rearrangement
b
1
, b
2
, ···b
n
1
, c
1
, b
n
1
+1
, ···b
n
2
, c
2
, b
n
2
+1
, ···b
n
3
, c
3
, ···
So the sum up to c
k
is at least k. So the partial sums tend to infinity.
We can prove an even stronger result:
Lemma. Let
P
a
n
be a series that converges absolutely. Then for any bijection
π : N → N,
∞
X
n=1
a
n
=
∞
X
n=1
a
π(n)
.
Proof.
Let
ε >
0. We know that both
P
|a
n
|
and
P
|a
π(n)
|
converge. So let
M
be such that
P
n>M
|a
n
| <
ε
2
and
P
n>M
|a
π(n)
| <
ε
2
.
Now N be large enough such that
{1, ··· , M} ⊆ {π(1), ··· , π(N)},
and
{π(1), ··· , π(M)} ⊆ {1, ··· , N}.
Then for every K ≥ N ,
K
X
n=1
a
n
−
K
X
n=1
a
π(n)
≤
K
X
n=M+1
|a
n
| +
K
X
n=M+1
|a
π(n)
| <
ε
2
+
ε
2
= ε.
We have the first inequality since given our choice of
M
and
N
, the first
M
terms of the
P
a
n
and
P
a
π(n)
sums are cancelled by some term in the huge
sum.
So
∀K ≥ N
, the partial sums up to
K
differ by at most
ε
. So
|
P
a
n
−
P
a
π(n)
| ≤ ε.
Since this is true for all ε, we must have
P
a
n
=
P
a
π(n)
.