Sequences and Series 201At any point zo E D where the series of the absolute values of the
terms, namely,
00
L lun(z)I = luo(z)I + · · · + lun(z)I + · · · (4.10-8)
n=Ois convergent, the series (4.10-6) also converges, by virtue of Theorem 4.6.
Thus the set A of all those points for which (4.10-8) converges is a subset
of C, and it is called the absolute convergence set of (4.10-6).
The Cauchy condition for uniform convergence on a set B, as applied
to series, becomes
ISm(z) - Sn(z)I = lun+1(z) + · · · + um(z)I < E
form > n > N. and all z E B. Letting m = n + p, with p > 0 an arbitrary
integer, we have, alternatively,
lun+i(z) + · · · + Un+p(z)I < E (4.10-9)
for n > N. and all z E B.
The following criterion, called the Weierstrass M -test for absolute and
uniform convergence, is very useful in practice. It gives a set E C A n B
where the series is both absolutely and uniformly convergent, not neces-
sarily the largest set of absolute and/or uniform convergence. In fact, in
most cases of interest, a series of functions has sets of uniform convergence,
but there is not a largest set where the series converges uniformly (see the
example below).
Theorem 4.17 Let E:;=O un(z), z E D, be a series of functions and
E:;=O Mn a numerical series of real positive terms. Suppose that:
l. lun(z)I S kMn for all n and all z EEC D, k being a positive constant.
- The series L Mn is convergent.
Then L un(z )converges absolutely and uniformly on E.
Proof Since L Mn converges, to every E > 0 there corresponds an N. such
that for n > N. and arbitrary p > 0, we have
tMn + ... + Mn+p < k
Hence, by using assumption (1), we obtain
lun(z) + · · · + Un+p(z)I S lun(z)I + ·: · + lun+pl(z)
S k(Mn + "· + Mn+p) < E