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multiplication of series∗
pahio†
2013-03-21 19:45:08
P∞
P∞
Theorem. If the series k=1 ak and k=1 bk with real or complex terms converge and have the sums A and B, respectively, and at least one of them converges absolutely, then also the series
a1 b1 +(a1 b2 +a2 b1 )+(a1 b3 +a2 b2 +a3 b1 )+ · · ·
(1)
is convergent and its sum is equal to AB.
Proof. Denote the partial sums of the series An := a1 + a2 + · · · + an ,
Bn := b1 + b2 + · · · + bn and sn := a1 b1 + (a1 b2 + a2 b1 ) + (a1 b3 + a2 b2 + a3 b1 ) +
· · · + (a1 bn + a2 bn−1 + · · · + an b1 ) for each n. ThenP
we have limn→∞ An = A
and limn→∞ Bn = B. Suppose that e.g. the series
an converges
absolutely
P∞
and that at least one an is distinct from zero; so the sum
n=1 |an | is a real
positive number M . Let ε be an arbitrary positive number.
Now we can write the identities
AB = (A − An )B + a1 B + a2 B + · · · + an B,
sn = a1 (b1 + b2 + · · · + bn ) + a2 (b1 + b2 + · · · + bn−1 ) + · · · + an b1 = a1 Bn +
a2 Bn−1 + · · · + an B1 ,
AB −sn = (A − An )B + a1 (B − Bn ) + a2 (B − Bn−1 ) + · · · + an (B − B1 )
= (A − An )B + [a1 (B − Bn ) + a2 (B − Bn−1 ) + · · · + ak (B − Bn−k+1 )]
+ ak+1 (B − Bn−k ) + · · · + an (B − B1 ).
ε
There is a positive number n1 (ε) such that |A−An | < 3(|B|+1)
when n >
n1 (ε). Then
|(A−An )B| = |A−An | · |B| <
ε
ε
(|B|+1) = .
3(|B|+1)
3
(2)
P
The convergence of
bn implies that there is a number n2 (ε) such that |B −
ε
Bn | < 3M
when n > n2 (ε). Thus we have
|[. . .]| ≤ |a1 |·|B −Bn |+ · · · +|ak |·|B −Bn−k+1 | < (|a1 |+ · · · +|ak |)
∗ hMultiplicationOfSeriesi
ε
ε
ε
≤ M·
=
3M
3M
3
(3)
created: h2013-03-21i by: hpahioi version: h37431i Privacy
setting: h1i hTheoremi h40A05i
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1
if n−k+1 > n2 (ε). Because limn→∞ Bn = B, the numbers |Bn | are bounded,
i.e. there is a positive number K such that for each j we have |Bj | < K and
consequently |B| ≤ K. It follows that |B − Bj | ≤ |B| + |Bj | < K +P
K = 2K
∞
for every j. We apply Cauchy criterion for convergence to the series n=1 |an |
getting a number n3 (ε) such that for each m, one has the inequality |ak+1 | +
ε
if k > n3 (ε). Accordingly we obtain the estimation
· · · + |am | < 6K
|ak+1 (B −Bn−k )+ · · · +an (B −B1 )| ≤ |ak+1 ||B −Bn−k |+ · · · +|an ||B −B1 | < 2K ·
(4)
which is valid when k > n3 (ε).
If we choose n > max{n1 (ε}, n2 (ε) + n3 (ε)} and k such that n ≥ k >
n3 (ε)+1, then the inequalities (2), (3) and (4) are satisfied, ensuring that
|AB −sn | <
ε ε ε
+ + = ε.
3 3 3
This means that the assertion of the theorem has been proved.
Remark. The mere convergence of both series does not suffice for convergence of (1). This is seen in the following example where both series are
1
1
1− √ + √ − + · · ·
2
3
They converge by virtue of Leibniz test, but not absolutely (see the p-test). In
their product series
1
1
1
1 1
1
1−( √ + √ )+( √ + √ √ + √ )− + · · ·
2
2
3
2 2
3
1
the absolute value of the nth term is 1· √1n + √12 √n−1
+ · · · + √1n · 1, having n
2
summands which all are greater than n+1 (this is seen when one looks at the
√ √
n+1 2
2
2
half circle y = x n+1−x or (x − n+1
2 ) + y = ( 2 ) , which shows that
n+1
1
2
2
y ≤ 2 and thus y ≥ n+1 ). Because limn→∞ n · n+1
= 2 6= 0, the product
series does not satisfy the necessary condition of convergence and therefore the
series diverges.
References
[1] Ernst Lindelöf: Johdatus funktioteoriaan. Mercatorin Kirjapaino Osakeyhtiö. Helsinki (1936).
2
ε
ε
=
6K
3