6
Series
We call a normed space (X, k·k) a Banach space provided that every Cauchy
sequence (xn ) in X converges. For example, R with the norm k·k = |·| is an
example of Banach space. Now let (xn ) be a sequence in X. Define a new
sequence (sn ) in X by
n
X
xk , n ∈ N.
sn =
k=1
P
P
The sequence (sn ) is called a series in X and is written as
xk or k xk .
The n-th term sn of the sequence (sn) isPcalled the nth partial sum and xk
is called the kth summand of the series k xk .
P
Definition 6.1. The series k xk converges or is convergent if the sequence
(sn ) of partial
limit limn→∞ sn is
P sums converges. TheP
Pcalled the value of
∞
the series k xk and is written as k=1 xk Finally,
k xk diverges or is
divergent if the sequence (sn ) diverges in X.
Proposition 6.2 (Necessary condition for convergence). If the series
P
xk converges, then xn → 0.
P
Proof. Denoting by sn the nth partial sum of the series
xk , we have
sn → x for some x. Since
xn = sn+1 − sn → x − x = 0,
the result follows.
Example 6.3. The series
e.
P
P
(1/k!) converges in R. Its value is ∞
k=0 1/k! =
P
P
Example 6.4. The series 1/k2 converges in R. Indeed, if sn = nk=1 1/k2 ,
then
n n
n
X
X
X
1
1
1
1
2
=1+
−
= 1 + 1 − < 2.
1/k ≤ 1 +
sn =
k(k − 1)
k−1 k
n
k=1
k=2
k=2
So, the sequence (sn ) of partial sums is bounded in R and since it is also
increasing, it converges.
P
P
Example 6.5. The harmonic series 1/k diverges. Indeed, if sn = nk=1 1/k,
then
1
n
1
1
+ ... +
≥
=
|s2n − sn | =
n+1
2n
2n
2
P
showing that (sn ) is not a Cauchy sequence and this implies that k 1/k is
dirvergent.
30
P k
Example 6.6. The geometric series
a , where a ∈ R satisfies |a| < 1,
converges. Indeed,
n
X
ak+1 − 1
ak =
sn =
a−1
k=0
and since |a| < 1, we have
ak+1
→ 0 showing that
lim sn =
n→∞
P
∞
X
ak =
k=0
1
.
1−a
ak diverges.
P
P
Proposition 6.7. Let
ak and
bk be convergent series in a Banach
space X and let α ∈ R. Then
P
P∞
P∞
• The
(ak + bk ) converges and
k=1 (ak + bk ) =
k=1 ak +
P∞ series
b
.
k=1 k
P
P∞
P∞
• The series ∞
k=1 (αak ) converges and
k=1 (αak ) = α
k=1 ak .
P
Theorem 6.8 (Cauchy criterion). For a series
xk in a Banach space
(X, k·k) the following conditions are equaivalent.
P
(a) The series
xk converges.
If |a| ≥ 1, then
k
(b) For every ε > 0 there is N ∈ N such that
m
X
xk < ε
k=n+1
for all m > n ≥ N .
P
P
Proof. Let sn = nk=1 xk be the nth partial sum. Then sm −sn = m
k=n+1 xk
if m > n.Thus (sn ) is Cauchy in X if and only if the condition (b) holds
true. By assumption, (X, k·k) is a Banach space and the result follows. P
P
Theorem 6.9. If
ak is a series in R and ak ≥ 0 for all k ∈ N, then
ak
converges if and only if the sequence (sn ) of partial sums is bounded. In this
case, the series has the value equal to supn∈N sn .
Proof. Since ak ≥ 0 for all k ∈ N, the sequence (sn ) of partial sums is
increasing. Hence (sn ) converges if and only if (sn ) is bounded.
31
6.1
Alternating series
P
Definition 6.10. A series
ak in R is called alternatingP
if ak and ak+1
have opposite signs. Any such series can be written as ± (−1)k ak with
ak ≥ 0 for all k ∈ N.
Theorem 6.11 (Leibniz criterion). Let (ak ) be a decreasing null
Psequence
such that ak ≥ 0 for all k ∈ N. Then the alternating series
(−1)k ak
converges in R.
Proof. Since
s2n+2 − s2n = −a2n+1 + a2n+2 ≤ 0,
the sequence (s2n ) is decreasing. Similarly,
s2n+3 − s2n+1 = a2n+2 − a2n+3 ≥ 0,
implying that (s2n+1 ) is increasing. Moreover, s2n+1 ≤ s2n so that s2n+1 ≤
s2 and s2n ≥ 0 for all n ≥ 1. Hence, there are s and t such that s2n → s
and s2n+1 → t. Also
t − s = lim (s2n+1 − s2n ) = lim a2n+1 = 0
n→∞
n→∞
implying that t = s. Take ε > 0. Then there are N1 , N2 ∈ N such that
|s2n − s| < ε,
2n ≥ N1
and
|s2n+1 − s| < ε,
2n + 1 ≥ N2 .
Thus, |sn − s| < ε for all n > 2 max{N1 , N2 }.
Example 6.12. The assumption that (an ) is decreasing is necessary. For
example, set
1
1
and a2n−1 =
.
a2n =
2n
(2n − 1)2
1
1
1
Then a2n−1 = (2n−1)
2 ≤ 2n−1 so that 0 ≤ an ≤ n showing that an → 0.
Since a2n−1
P< a2nn for n ≥ 2, the sequence (an ) is not decreasing. However,
the series (−1) an diverges. Arguing by contradiction we assume that the
sequence (sn ) of partial sums converges. We have
s2n =
2n
X
(−1)k ak =
k=1
n
X
=−
k=1
n
n
X
X
(−1)a2k−1 +
a2k
k=1
1
+
(2k − 1)2
= −An + Bn ,
32
k=1
n
X
k=1
1
2k
P
Pn 1
1
1
1 Pn
where An = − nk=1 (2k−1)
2 and Bn =
k=1 k . Since the
k=1 2k = 2
P
1
series k≥1 (2k−1)
2 converges, the sequence (An ) converges. Consequently,
the sequence
B
=
sn + An converges. But this contradicts the fact that the
n
P
1
series k≥1 k diverges.
6.2
Absolute convergence
P
Definition 6.13. The series
xk in a BanachPspace (X, k·k) converges
absolutely or
Pis absolutely convergent if the seriesP kxk k is convergent
P in R.
The series
xk is conditionally convergent if
xk converges but
kxk k
diverges.
P
Example 6.14. The alternating harmonic series (−1)k /k converges
in
P
view of Leibniz criterion,Phowever, the series of absolute values
1/k diverges. Hence the series (−1)k /k converges conditionally.
Proposition 6.15. Every absolutely convergent series converges.
Proof. The proposition is a consequence of Cauchy criterion and the triangle
inequality. Indeed,
m
m
X
X
xk ≤
kxk k .
k=n+1
k=n+1
6.3
Comparison, Root, and Ratio Tests
From now on we only consider real sequences, however, you are invited to
figure out which of the results can be stated in the full generality of Banach
spaces.
P
P
Proposition 6.16 (Comparison test). Let
ak and
bn be two sequences of nonnegative terms and assume that
an ≤ bn
for all n.
Then the following holds true,
P
P
(a) If
bk converges, then so is
ak .
P
P
(b) If
ak diverges, then so is
bk .
33
P
m
Example 6.17. Let m ≥ 2. Then the series
P 1 1/k converges absolutely.
Recall from Example 6.4 that the series
converges. Since for m ≥ 2,
k2
we have
1
1
≤ 2
for all k,
m
k
k
P
it follows from Proposition 6.16 that the series 1/km is indeed convergent.
P
P
Proposition 6.18 (Limit comparison test). Let
ak and
bn be two
sequences of positive terms and assume that
an
→L
bn
(Clearly, L ∈ [0, ∞) or L = ∞). Then the following holds true,
P
(a) P
If L ∈ (0, ∞), then the convergence
of
bk implies the convergence
of
P
P
ak and the divergence of
ak implies the divergence of
bk .
P
P
(b) Assume that L = 0. Then
ak converges if
bk converges.
P
P
(c) Assume that L = ∞. Then
ak diverges, if
bk diverges.
Proof. We only prove (a). If L ∈ (0, ∞), then there is N so that L/2 =
L−L/2 < an /bn < L+L/2 = 3L/2 for n ≥ N . So, (L/2)bn < an < (3L/2)bn
for n ≥ N and the result follows from Proposition 6.16.
P
Theorem 6.19 (Root test). Let
ak be a series of nonnegative real numbers and let
p
α = lim sup n |an |.
Then
•
•
P
P
ak converges absolutely if α < 1.
ak diverges if α > 1.
P
If α = 1, then both convergence and divergence of
ak are possible.
P
Theorem 6.20 (Ratio test). Let
ak be a series of real numbers such
that ak 6= 0 for all k ≥ N . Then
• If
then
P
ak+1 < 1,
lim k→∞ ak ak converges absolutely.
34
• If
then
P
ak+1 > 1,
lim k→∞ ak ak diverges.
P
a If limk→∞ k+1
ak are
ak = 1, then both convergence and divergence of
possible.
Proposition 6.21 (Cauchy’s condensationPtest). Let (ak ) be a sequence
of nonnegative
ak converges if and only if
P k and decreasing terms. Then
the series
2 a2k converges.
P
P
Proof. Denote by sn = nk=1 and tn = nk=1 2k a2k . Then, if n < 2m , we
have
sn ≤ a1 + (a2 + a3 ) + . . . + (a2m + . . . + a2m+1 −1 )
≤ a1 + 2a2 + . . . + 2m a2m = tm .
P
If 2k a2k converges, then the sequence
P (tm ) is bounded and above estimate
shows that (sn ) is bounded. Hence
ak converges. For the converse, note
that if n > 2m , then
sn ≥ a1 + a2 + (a3 + a4 ) + . . . + (a2m−1 +1 + . . . + a2m )
1
1
≥ a1 + a2 + 2a4 + . . . + 2m−1 a2m = tm .
2
2
P
This P
shows that if
ak converges, then the sequence (tm ) is bounded so
k
that
2 a2k converges.
P 1
Example 6.22. Consider
series
the
np with p > 0.PThen
n
P condenP the
2n
1
1
converges
if
and
only
if
=
sation test implies that
p
np
p−1
n
2
P 2 1 n
converges
converges. We already know that the geometric series
2p−1
1
< 1, i.e., p > 1.
if and only if 2p−1
The next theorem is a generalization of the condensation test.
Theorem 6.23 (Schlömlich theorem). Let (gk ) be a strictly increasing
sequence of positive integers such that gk+1 − gk ≤ C(gk − gk−1 ) for some
C > 0 and all k ∈ N. If (ak ) is a decreasing sequence of nonnegative
numbers, then
X
X
ak converges if and only if
(gk+1 − gk )agk converges.
35
Proof. The proof is similar to the proof of the condensation
test. Let (Sn )
P
be the sequence of the partial sums
of
the
series
a
and
(T
)
the sequence
n
k
P
of the partial sums of the series (gk+1 − gk )agk . Now, if n < gk , then
Sn ≤ Sg k
≤ (a1 + . . . + ag1 −1 ) + (ag1 + . . . + ag2 −1 ) + . . . + (agk + . . . + agk+1 −1 )
≤ (a1 + . . . + ag1 −1 ) + (g2 − g1 )ag1 + . . . + (gk+1 − gk )agk
≤ (a1 + . . . + ag1 −1 ) + Tk .
Consequently, if the sequence (Tk ) converges
and hence is bounded, then
P
the sequence
(S
)
is
bounded
so
that
a
converges.
Conversely, assume
n
k
P
that
ak converges. If n > gk , then
CSn ≥ CSgk
≥ C(ag1 +1 + . . . + ag2 ) + . . . + C(agk−1+1 + . . . + agk
≥ C(g2 − g1 )ag2 + . . . + C(gk − gk−1 )agk
≥ (g3 − g2 )ag2 + . . . + (gk+1 − gk )agk = Tk − (g2 − g1 )ag1 .
P
Consequently, the sequence (Tk ) is bounded and the series (gk+1 − gk )agk
converges.
Example 6.24. Take gk = 3k . Then the sequence (gk ) satisfies the assumptions of the above theorem and gk+1 − gk = 3k+1 − 3k = 2 · 3k . So, the
following holds true. Assume that (ak ) is a decreasing sequence of nonnegative numbers, then
X
X
ak converges if and only if
3k a3k converges.
P 1
Now consider the series
. Since the sequence ( 3ln1 n ) is decreasing
P 3n
P 1 3ln n
P 3n
and converges to 0,
=
ln n converges if and only if
ln 3n =
3
3n ln 3
3
n
P
1
ln
3−1
converges. Since ln 3 > 1, it follows that 3
> 1 and
3ln 3−1
n
P
P 1
1
1
< 1 so that the series
conconverges. Hence
3ln 3−1
3ln 3−1
3ln n
verges.
Consider gk = k2 , then the sequence (gk ) satisfies the assumptions of the
theorem. Then the following is true. If (ak ) is a decreasing sequence of
nonnegative numbers, then
X
X
ak converges if and only if
kak2 converges.
P
Indeed, since (k + 1)2 −P
k2 ) = 2k + 1, it follows from Theorem ?? that
ak
converges if and only if (2k + 1)ak2 converges. Since kak2 ≤ (2k + 1)ak2 ≤
36
P
if and only if
3kak2 , the comparison test shows that (2k + 1)ak2 converges P
P
1
√
kak2 converges. showing our claim. Now consider the series
. Since
2 nP
√
1
n
√
the sequence 1/2 ) is decreasing P
and converges
to 0, the series
2 n
P
n
√n
=
converges if and only if the series
2n converges. Using the
2 n2
P n
P 1
√
ratio test, we see that
converges.
2n converges and so
2 n
6.4
The Dirichlet and Abel Test
We start with the following lemma.
Lemma 6.25 (Abels’ lemma). Let (xn ) and (yn ) be two sequences of real
numbers. Let (sn ) be a sequence of partial sums of (yn ) and s0 = 0. Then
m
X
k=n+1
xk yk = (xm sm − xn+1 sn ) +
m−1
X
(xk − xk+1 )sk .
k=n+1
Proof. Note that yk = sk − sk−1 . So,
m
X
k=n+1
xk yk =
m
X
k=n+1
xk (sk − sk−1 )
= xn+1 (sn+1 − sn ) + xn+2 (sn+2 − sn+1 ) + . . . + xm (sm − sm−1 )
= xm (xm − xn+1 sn ) + sn+1 (xn+1 − xn+2 ) + . . . + sm−1 (xm−1 − xm )
= xm (xm − xn+1 sn ) +
m−1
X
k=n+1
xk (sk − sk−1 )
We use this lemma to obtain a test for convergence of the series
P
xk yk .
Theorem 6.26 (Dirichlet’s test). Suppose that (xn ) is a decreasing sequence such that xn → 0 and
Pthe sequence (sn ) of the partial sums of (yn )
is bounded. Then the series
xk yk converges.
Proof. Since (sn ) is bounded, |sn | ≤ C for all n ≥ 1. Since (xk ) is decreasing,
xk − xk+1 ≥ 0. Thus, by the above lemma,
m
m−1
X
X
≤
|x
s
|
+
|x
s
|
+
x
y
(xk − xk+1 ) |sk |
m m
n+1 n
k k
k=n+1
k=n+1
≤ ((xn+1 + xm ) + (xn+1 − xm ))C = 2xn+1 C.
37
P
Example 6.27. Consider (−1)n xn where xn is a decreasing sequence of
nonnegative numbers converging to 0. Then the series converges in view
of alternating series test. However,Pthis also follows
from Dirichlet’s test.
Pn
n
n
Indeed, let yn = (−1) . Then sn = k=1 bk = k=1 (−1)k is equal to −1 if
n is odd and is equal to 0 if n is even. Hence
(sn ) is bounded,
P the sequence
n
and Dirichlet’s test implies that the series (−1) an converges.
Example 6.28. One can show, using for example the principle of mathematical induction, that
sin
ϑ
(n + 1)ϑ
nϑ
· (cos ϑ + cos 2ϑ + . . . + cos nϑ) = cos
· sin
.
2
2
2
Hence if ϑ 6= 2πk for all k ∈ Z, then
(n+1)ϑ
nϑ ·
sin
cos 2
2 1
.
≤
|cos ϑ + cos 2ϑ + . . . + cos nϑ| =
ϑ
sin sin ϑ 2
2
Similarly,
sin
ϑ
(n + 1)ϑ
nϑ
· (sin ϑ + sin 2ϑ + . . . + sin nϑ) = sin
· sin
2
2
2
which implies that
1
|sin ϑ + sin 2ϑ + . . . + sin nϑ| ≤ sin ϑ 2
if ϑ 6= 2πk for all k ∈ Z. Hence, in view of Dirichlet’s test, one has the
following result.
Corollary 6.29. If (xn ) is a decreasing
sequence converging
to 0 and ϑ 6=
P
P
2πk for all k ∈ Z, then the series
xn cos nϑ and
xn sin nϑ converge.
Theorem 6.30P(Abel’s test). If (xP
n ) is a monotone convergent sequence
and the series
yk converges, then
xk yk converges.
Proof. Without loss of generality, we may assume that (xn ) is decreasing.
If x = lim xn , set un = xn − x is decreasing and converging to 0. P
Thus,
xn = x + un and
x
y
=
xy
+
u
y
.
Now
note
that
the
series
xyk
n
n n
P n n
P
converges since
yk converges and the
un yn converges in view of
P series
the Dirichlet’s test. Hence, the series
xn yn converges.
38
6.5
Cauchy Product
Definition 6.31.
The Cauchy product of two series
P
is the series n≥0 cn where
cn =
n
X
ak bn−k .
k=0
P
n≥0 an
and
P
n≥0 bn
n ≥ 0.
P
Example 6.32. Let
n≥0 cn be the Cauchy product of the alternating
P (−1)n
√
series
with itself. That is,
n+1
n
n
X
X
1
(−1)n−k
(−1)k
n
√
p
= (−1)
.
·p
cn =
k+1
(n − k) + 1
(k + 1)(n − k)
k=0
k=0
P (−1)n
√
Note that , in view of the alternating series test, the series
converges
n+1
conditionally. However, the Cauchy product diverges. To see this we show
that the sequence (cn ) does not converge to 0. Indeed, we have
(n − k + 1)(k + 1) =
2 n
2
+1 −
−k
2
2
n
and hence
q
p
p
n
(n − k + 1)(k + 1) = (n/2 + 1)2 − (n/2 − k)2 ≤ n/2 + 1)2 = + 1.
2
Using this we find that
|cn | =
n
X
k=0
p
1
(k + 1)(n − k)
≥
n
X
k=0
2
1
= (n + 1)
≥1
n/2 + 1
n+2
showing that the sequence (cn ) does not converge to 0.
P
Theorem
6.33
(Mertens
theorem).
Assume
that
n≥0 converges absoP∞
P∞
P
lutely,
Cauchy product of
n = A,
n = B, and
n=0 aP
n=0 bP
n≥0 is the
P
P∞
a
and
b
.
Then
c
converges
and
n
n
n
n≥0
n≥0
n≥0
n=0 cn = AB.
Proof. Define
An =
n
X
k=0
an ,
Bn =
n
X
bn ,
Cn =
k=0
n
X
k=0
39
cn ,
βn = Bn − B.
Then
Cn = c0 + c1 + . . . + cn
= a0 b0 + (a0 b1 + a1 b0 ) + . . . + (a0 bn + a1 bn−1 + . . . + an b0 )
= a0 (b0 + b1 + . . . + bn ) + a1 (b0 + b1 + . . . + bn−1 ) + . . . + an b0
= a0 Bn + a1 Bn−1 + . . . + an B0
= a0 (βn + B) + a1 (βn−1 + B) + . . . + an (β0 + B)
= (a0 + . . . + an )B + (a0 βn + a1 βn−1 + . . . + an βn )
= An B + γn
where we have abbreviated
γn = a0 βn + a1 βn−1 + . . . + an βn .
Since An B → AB, it suffices to show that γn → 0. In order to show that
γn → 0, pick ε > 0. Then, since βn → 0, there is N such that
Let
(The series
Then
P
b=
A
n≥0 |an |
|βn | < ε
for all n ≥ N .
∞
X
and
n=0
|an |
|βn | ≤ M.
converges and (βn ) is bounded since it converges.)
|γn | = |a0 βn + a1 βn−1 + . . . + an β0 |
≤ |a0 βn + a1 βn−1 + . . . + an−N βN |
+ |an−N +1 βN −1 + an−N +2 βN −2 + . . . + an β0 |
b + M (|an−N +1 | + |an−N +2 | + . . . + |an |)
≤ε·A
Keeping N fixed and letting n → ∞, we get
b
lim sup |γn | ≤ εA
since an → 0. Since ε > 0 was arbitrary, we conclude that
lim sup |γn | = 0
which implies that lim γn = 0 as claimed
P
P∞
Theorem 6.34 (Abel’s theorem).
If ∞
n=0 an = A,
n=0 bn = B, and
P
P
n
∞
a
b
for
all
n
≥
0,
then
C = AB.
c
=
C,
where
c
=
n
n
k
n−k
k=0
n=0
40
6.6
Double Sums
Let (X, k·k) be a normed
P space, let (aij )(i,j)∈N×N be a double sequence in X,
and the double series aij . We have a function a : N×N → N, (j, k) 7→ ajk .
Since N × N is countable,
there is a bijection σ : N → N × N. If σ is
P
Psuch
bijection, the series n aσ(n) is called an ordering
of
the
double
series
ajk .
P
If we fix j ∈ N (or k ∈ N ), then the series
ajk is called the jth row series
(or the kth column series). If every row seriesP(orP
column series) converges,
∞
then we may consider
the
series
of
row
sums
(
j
k=0 ajk
P P∞
P) (or the series of
column sums k ( j=0 ajk )). Finally, the double series
ajk is summable
if
n
X
sup
|ajk | < ∞.
n
Theorem 6.35. Let
j,k
P
ajk be a summable double series. Then
P
P
(a) Every ordering
ajk is converges absolutely to a value
n aσ(n) of
s ∈ X which is independent of σ.
P P
(b) P
The P
series of row sums j ( ∞
k=0 ajk ) and the series of column sums
∞
(
a
)
converge
absolutely
and
k
j=0 jk
∞
∞ X
∞
X
XX
(
ajk ) =
ajk ).
(
j=0 k=0
k
j=0
P
To prove (a) we will need the following result. Let
an be a series
P and
let σ : N → N be a bijection.
Define
b
=
a
.
Then
the
series
bn is
n
σ(n)
P
called a rearrangement of of
an .
P
Proposition 6.36.
Assume
that
an converges absolutely. Then any reP
arrangement of
an converges absolutely. (They all converge to the same
value).
P
Proof. Assume
an converges absolutely. Take a bijection σ : N → N
Pthat
and consider
bn P
with bn =P
aσ(n) . Denote by (An ) and
P(Bn ) the sequences
of partial sums of
an and
bn , respectively. Since
an converges absolutely, given ε > 0, there is N such that
m
X
k=n
|ak | < ε
41
(1)
for all m > n ≥ N . Let K be such that {1, . . . , N } ⊂ {σ(1), . . . , σ(K)}.
Take n > K and consider the difference |An − Bn | Note that the terms
a1 , . . . , aN appear in both partial sums An and Bn . So, by (1)
|An − Bn | < ε.
P
Hence lim An = lim Bn , proving that
bn converges to the same value
P
P as
anP
. The same argument in which P
An and Bn are partial sums of
|an |
and
|bn |, respectively, shows that
bn converges absolutely.
P P
Proof of Theorem 6.35. (a) Abbreviate M = n nj,k=0 |ajk |. Let σ :
N → N × N be a bijection and let N ∈ N. Then there is some K ∈ N such
that
{σ(0), . . . , σ(N )} ⊂ {(0, 0), (1, 0), . . . , (K, 0), . . . , (K, K)}.
(2)
P
Together with the summability of
ajk this imples
N
K
X
X
aσ(n) ≤
|ajk | ≤ M.
n=0
j,k=0
P
This implies that n aσ(n) converges absolutely. Let δ : N → N × N be
another bijection. Then α = σ −1 ◦ δ : N → N is a bijection. Set ym = xσ(m) .
Then
bα(n) = aσ(α(n)) = aδ(n) , n ∈ N.
P
P
P
a
.
Since
of
Thus n aδ(n) is a rearrangement
σ(n)
n aσ(n) converges
n
P
absolutely, the series P n aδ(n) also converges absolutely.
P∞
∞
(b) The row series
j=0
k=0 ajk , j ∈ N, and the column series
P ajk ,
k ∈ N, converge absolutely. PThis
follows from the summability of
ajk .
P∞
a
)
and
the
series
of
column
series
(
So,
the
series
of
row
series
jk
k=0
j
P P∞
k ( j=0 ajk ). To see that these series converge absolutely, consider the
following inequalities,
m
m
m
l X
l X
X
X
X
|ajk | ≤ M, l ≤ m.
|ajk | ≤
ajk ≤
j=0 k=0
j=0 k=0
j,k
Pl
P∞
Taking the limit m → ∞, we get
l ∈PN. This
j=0 | k=0 ajk | ≤ M ,P
proves the absolute convergence of the series of row series j ( ∞
k=0 ajk ).
A similar
proves the absolute convergence of the series of column
P argument
P
series k ( ∞
a
).
j=0 jk
42
P∞
Now let σ : N → N × N be a bijection
s = n aσ(n) . Then for
P∞and let
every ε > 0, there is N ∈ N such that n=N +1 aσ(n) < ∞. Also there is
some K ∈ N such that (2) is satisfied. Then
X
m
∞
N
X
X
l X
aσ(n) < ε/2, l, m ≥ K.
a
≤
a
−
jk
σ(n)
j=0 k=0
n=N +1
n=0
Taking the limit m → ∞ and l → ∞, we get
X
∞
∞
N
∞ X
X
X
aσ(n) ≤ ε/2.
≤
a
a
−
jk
σ(n)
n=N +1
j=0 k=0
n=0
Applying the triangle inequality to
∞
N
X
X
aσ(n) < ε/2,
aσ(n) ≤
s −
n=0
n=N +1
we get
∞ ∞
X X
ajk − s ≤ ε.
j=0 k=0
P P
A similar argument shows that the value of k ( ∞
j=0 ajk ) is also s.
43