1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences and Series

then also

1. Im Sn - = L

n--+oo Pn

177

(Stolz)

Proofs The proofs of properties 1 to 5 follow easily from Definition 4.3 and

4.4 and are left to the reader. The proofs of properties 6 to 12 do not differ

essentially from those for the corresponding properties in Theorem 3.2,

and property 14 is a particular case of Theorem 3.12. We proceed to give

proofs of properties 13, 15, and 16.

(13) Suppose that the sequence {f (zn)} converges for every sequence

{zn} such that Zn_, a with Zn ED and Zn-/= a. Then {f(zn)} always con-

verges to the same limit, irrespective of the particular {Zn} chosen. In fact,

suppose that f(z~) _, L 1 and f(z~) _, L 2 , where {z~} and {z~} satisfy

the required conditions. Consider a new sequence {zn} where Z2n-1 = z~

an d Z2n = Zn II (' i.e., cons1 'd er th e sequence zI 1 , z 1 II , zI 2 , z 2 II ,.... ) Th' is new

sequence is also of the required type, so {f(zn)} converges to some limit

L. But {f(z~)} and {f(z~)} are subsequences of the convergent sequence

{f(zn)}, and hence have the same limit by property 3, so that L 1 = L 2 = L.

Now assume that L is not the limit of f at a (note that nothing is sup-

posed concerning the existence of limz--+a f(z)). Then there is some € > 0

such that for every 5 > 0 there is ( E D with 0 < I( - al < 5, for which

If ( () - LI ~ €. In particular, for each positive integer n we may take

On = l/n, and then there is (n E D, with 0 < l(n - al < l/n, for which

lf((n) - LI ~ €. Thus we have a sequence {(n} such that (n _, a, with

(n E D - {a}, yet f ( (n) does not converge .to L. This contradiction shows

that f ( z) has a limit at a, and that this limit is L.

Next, suppose that limz--+a f ( z) = L, and let Zn _, a with Zn E D - {a}.

For every e > 0 there is a 5 > 0 such that 0 < lz - al < 5, z E D, implies

that lf(z) - LI < €. Since Zn _, a, there exists N such that for n > N

we have 0 < lzn - al < 5 and Zn ED. Hence lf(zn) - LI < €for n > N,

so {f(zn)} converges to L.
(15) Let Zn , L, and suppose that L = 0. Then we must show that
Wn , 0. For a given€> 0 there is an N such that lznl < e/2k for n > N.
Hence for n > N (N fixed), we have

lwnl < IP1llz1I + · · · + IPNllzNI + (IPN+1I + · · · + 1Pnl)(e/2k)

• IP1 + ' ' ' + Pn I IP1 + ' ' ' + Pn I
  IP1llz1I + · · · + IPNllzNI €
  < +-

- IP1 + " · + Pn I 2

since the assumption implies IP N +i I + · · · + IPn I ::; k IP1 + · · · + Pn I· Now

let M = max(lz 1 1, ... , lzNI). Then the first fraction on the right-hand side