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convergence of complex term series∗
pahio†
2013-03-22 2:20:38
A series
∞
X
cν = c1 +c2 +c3 + . . .
(1)
ν=1
complex terms
(aν , bν ∈ R ∀ ν)
cν = aν +ibν
is convergent iff the sequence of its partial sums converges to a complex number.
Theorem 1. The series (1) converges iff the series
∞
X
aν
and
ν=1
∞
X
bν
(2)
ν=1
formed by real parts and the imaginary parts of its terms both are convergent.
Proof. Let ε > 0. Denote
n
X
ν=1
aν := sn ,
n
X
bν := tn ,
ν=1
n
X
cν := un .
ν=1
If the series (2) are convergent with sums S and T , then there is a number N
such that
ε
ε
|sn − S| < , |tn − T | <
when n = N.
2
2
Consequently,
p
|un −(S+iT )| = (sn − S)2 + (tn − T )2 5 |sn −S|+|tn −T | < ε when n = N,
i.e. the series (1) converges to S+iT . If, conversely, (1) converges to a complex
number
u = s+it (s, t ∈ R),
∗ hConvergenceOfComplexTermSeriesi
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then
|sn −s| 5 |(sn −s)+i(tn −t)| = |un −u|,
|tn −t| 5 |(sn −s)+i(tn −t)| = |un −u|,
and consequently, lim sn = s and lim tn = t, i.e. the series (2) are convern→∞
n→∞
gent with sums the real numbers s and t.
Theorem 2. The series (1) converges absolutely iff the series (2) both
converge absolutely.
Proof. The absolute convergence of (1) means that the series
∞
X
|cν |
ν=1
converges. But since |cν |2 = |aν |2 + |bν |2 , we have
|aν | 5 |cν |;
|bν | 5 |cν | 5 |aν | + |cν |.
From these inequalities we can infer the assertion of the theorem 2.
References
[1] R. Nevanlinna & V. Paatero: Funktioteoria. Kustannusosakeyhtiö
Otava. Helsinki (1963).
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