An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 83.

A convergent infinite series of powers is continuous.

139

since the moduli of the terms cannot increase, and only for n = co can
we have R Lim -+1 = 1. If then we give B a smaller value, the
An
property of absolute convergence is satisfied. Accordingly, series can be
semiconvergent, if at all, only in points on the circle of convergence*).
Since each of its terms is unique, an infinite series of powers as
long as it converges, is a one-valued function of the complex variable,
that does not anywhere become infinite. This function is continuous,
i e. when s and s + 6 are complex values for which the series converges,
Lim mod [f (z - 6) - f (z)] = 0, for 6= 0.
To prove this, we separate the terms of the series f () into the groups:
cp (Z) == Ca + a, z + - * * an-1 n - 1
() -= az + + an+z+ +...an+kn+k +...;
inasmuch as from the nth term onwards, within the circle of convergence:
An+ R < Ana, An +22 < An 2, *' An I Rk < An al
where a denotes a proper fraction, we must therefore have:
mod g (z)< AnR i- '
Accordingly merely by choosing a lower limit for n, we are able to
make both mod p (s) and mod (z + 6), and therefore also:
mod [p (Z - 6) - P (z)]
less than an arbitrarily small quantity e. Now, since:
d ) - mod [f (z +_ ) - f (z)] -= mod I[ (z ) - + () + ( ) - i (z)]
< mod [(p (z +_ ) - -p (z)] + e
*) Examples:
1. The binomial series:
m n - m(m- 1) z2 rn (m- 1) (mn- 2)~
+ m'+ (1 2 z +' - 1 2 3 3
in which m is a real number and z complex, converges absolutely as long as mod [z]
is less than 1; it diverges if mod [z] > 1. For mod [z]= -1:
If m > 0, it also converges absolutely along the entire circle of convergence.
If m < 0 but > - 1, the series is semiconvergent, with the exception of the
point z=- 1, in which it diverges.
If in <- 1, the series diverges in all points of the circle of convergence.
All these results can be deduced from ~ 46. In investigating the case -- 1 < m < 0
put Pn e 1 + miZ + -.. n zn, multiply both sides of the equation by (1 + z)
and consider the limiting value for n = oo.
2. The logarithmic series:
z2 z3 Z t
Z — + - +...
2+3 4
converges absolutely for mod [z] < 1, it diverges for mod [z] > 1. On the circle
of convergence it is semiconvergent, except at the point - -1 at which it
diverges.