Interchanging the Order of Summation
Corollary (Interchanging the Order of Summation)
If
∞ X
∞
X
ajk < ∞
∞ X
∞
X
then
j=1 k=1
Remark. The hypothesis
ajk =
j=1 k=1
∞ X
∞
X
ajk
k=1 j=1
P∞ P∞ k=1 ajk < ∞ really means that
j=1
for each j ∈ IN,
∞
X
ajk = Mj < ∞
∞
X
and
Mj < ∞
j=1
k=1
The two double sums in the conclusion really mean
∞
∞ X
X
j=1 k=1
ajk = lim
n→∞
∞
n X
X
j=1
= lim lim
n→∞ m→∞
∞ X
∞
X
k=1 j=1
ajk = lim
m→∞
ajk
k=1
n X
m
X
= lim lim
m→∞ n→∞
ajk
j=1
n X
m
X
n→∞
j=1
lim
m→∞
m
X
ajk
ajk
k=1
ajk
j=1 k=1
m
∞
X X
k=1
= lim
n X
= lim
m→∞
m X
k=1
lim
n→∞
n
X
j=1
ajk
j=1 k=1
That all of these limits exist is part of the conclusion of the corollary.
This result is a corollary of the following theorem, which has already been proven in
class.
Theorem. Let X be a metric space, E ⊂ X and p ∈ E ′ , the set of limit points of E. Let
f : E → C and, for each n ∈ IN, fn : E → C and assume that
(H1) lim fn (t) = f (t) uniformly on E and
n→∞
(H2) for each n ∈ IN, lim fn (t) = An exists
t→p
Then
(a) lim An = A exists and
n→∞
(b) lim f (t) = A. That is, lim lim fn (t) = lim lim fn (t).
t→p
c Joel Feldman.
t→p n→∞
2008. All rights reserved.
n→∞ t→p
February 4, 2008
Interchanging the Order of Summation
1
n→∞
z
}|

t  f1 (t) f2 (t) f3 (t) · · ·
↓
↓
↓
···
↓

p
A1
A2
A3 · · ·
Proof of the Corollary:
1
= tm and define
m
Set X = IR, E =
fn (tm ) =
n X
m
X
ajk
{
f (t)
↓(b)
unif
−→
(a)
−→
A
1 1
1
1, 2 , 3 , · · · m
, · · · and p = 0. Write
f (tm ) =
j=1 k=1
∞ X
m
X
ajk
j=1 k=1
P∞
The infinite sum in the definition of f (tm ) converges by comparison with j=1 Mj .
m
P
ajk ≤ Mj for all m ∈ IN. So the Weierstrass
Verification of (H1): For each j ∈ IN, k=1
M –test implies that fn converges to f , uniformly on E.
∞
P
Verification of (H2): For each j ∈ IN,
ajk converges absolutely by the hypothesis that
k=1
∞ P
ajk = Mj < ∞. So
k=1
lim fn (tm ) = lim
m→∞
m→∞
n X
m
X
n
X
ajk =
j=1 k=1
j=1
lim
m→∞
m
X
ajk = An
k=1
exists.
So the theorem now tells is that
lim An = lim lim
n→∞
n→∞ m→∞
n X
m
X
and
lim f (tm ) = lim lim
m→∞
ajk
j=1 k=1
m→∞ n→∞
exist and are equal.
n X
m
X
ajk
j=1 k=1
Example. Here is an example which illustrates the need for the hypothesis that the double
sum converges absolutely. We choose

1
if j = k = 1


1
if k = j + 1
ajk =

−1
if
k =j−1

0
otherwise
c Joel Feldman.
2008. All rights reserved.
February 4, 2008
Interchanging the Order of Summation
2
This example is rigged to give the partial sums
Smn =
m X
n
X
ajk =
j=1 k=1
(
1
2
0
if m = n
if n > m
if n < m
Pictorially
ajk
j
↓
k
→
1
1
0 0 0
−1 0
1 0 0
0
−1 0 1 0
0
0 −1 0 1
..
..
.. ..
.
.
.
.
Smn
m
↓
···
···
···
..
.
n
1
0
0
0
..
.
↓
→
2
1
0
0
..
.
↓
2 ··· →
2 ··· →
2 ··· →
1
0
↓
ց
2
2
2
0
0 0 0 → 0
1
2
2
1
0
..
.
↓
↓
2
For any fixed n, Sm,n = 0 for all m > n and so converges to 0 as m → ∞. Hence
m X
n
X
lim lim
n→∞ m→∞
ajk = lim
n→∞
j=1 k=1
lim Sm,n
m→∞
= lim 0 = 0
n→∞
Similarly, for each fixed m, Sm,n = 2 for all n > m and so converges to 2 as n → ∞. Hence
lim lim
m→∞ n→∞
m X
n
X
ajk = lim
j=1 k=1
m→∞
lim Sm,n
n→∞
= lim 2 = 2
m→∞
And the sequence Sm,m = 1 converges to 1 as m → ∞. So
lim
m→∞
c Joel Feldman.
m
m X
X
j=1 k=1
2008. All rights reserved.
ajk = lim Sm,m = lim 1 = 1
m→∞
February 4, 2008
m→∞
Interchanging the Order of Summation
3