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III - Convergence: Continuous variables
The functions as being continuous, so is the zeta function for s > 1. It
possesses more spectacular properties, but this would take us too far aside.
Note, however, that the Riemann series does not converge normally on
the interval X =]1, +00[. Indeed, for all n, 1/ns < lin for all s > 1 and it is
clear (or will become so in Chap. IV) that 1/ns tends to lin when s ----+ 1 +0;
so here Ilunllx = lin, which rules out the relation (3).
Example 3. Every power series which converges in a disc of radius R > 0
converges normally on any disc of radius strictly smaller than R.
Let r < R. We have known since Chap. II that the given series
fez) = I:anz n converges absolutely for Izl = r; since lanznl ::; lanrnl = Vn
for Izl ::; r, normal convergence follows and, incidentally, so again does the
continuity of the sum; but we knew that already!
The hypothesis r < R is essential. Consider for example the geometric
series I: zn which converges for Izl < 1 to the function s(z) = 1/(1 - z). We
shall show that the series does not converge norma.lly, nor even uniformly, on
the disc I z I < 1. If it did one could find an n such that
Is(z) - sn(z)1 ::; 1
on all the disc Izl < 1. The polynomial sn(z) = 1 + z + ... + zn being, in
absolute value, majorised by n + 1 for Izl < 1, one would then have Is(z)1 ::;
n + 2, i.e. 11 - zl ~ 1/(n + 2) for Izl < 1; impossible on a neighbourhood of
z = 1. Since furthermore Is(z) - sn(z)1 = Izln+l III - zl, it is clear that, for
n given, the left hand side increases indefinitely as Izl tends to 1, so is not
even bounded on Izl < 1.
If one restricts oneself temporarily to real values of z and if one remarks
that every compact, so closed, interval contained in X =] - 1, +1[ is in fact
contained in an interval [-q, +q] with q < 1, one has thus obtained a series
of continuous functions which, without converging uniformly on the open interval X, converges uniformly on every compact interval contained in X. A
very frequent phenomenon, valid for every power series and which nevertheless suffices to assure the continuity of the sum of the series. It is in fact
continuous on [-1, 1[ because it is equal to 1/(1 - z); the following example
will provide a better reason in a more interesting case.
Example 4. For x real consider the series
(8.7)
L(x) = x - x 2 /2 + x 3 /3 - ... = 2) _l)n+1xn In
n~l
whose sum is the function 10g(1 + x), as we shall see (nO 16, example 1). It
converges absolutely for Ixl < 1 since its general term is, in absolute value,
majorised by that of the corresponding geometric series. It converges for x = 1
also, since then it is the alternating harmonic series I:( _1)n+l In. Finally, it
diverges for x = -1 (harmonic series). Let us show that its sum is continuous
on the interval] - 1,1] where it is defined.
III - Convergence: Continuous variables
The functions as being continuous, so is the zeta function for s > 1. It
possesses more spectacular properties, but this would take us too far aside.
Note, however, that the Riemann series does not converge normally on
the interval X =]1, +00[. Indeed, for all n, 1/ns < lin for all s > 1 and it is
clear (or will become so in Chap. IV) that 1/ns tends to lin when s ----+ 1 +0;
so here Ilunllx = lin, which rules out the relation (3).
Example 3. Every power series which converges in a disc of radius R > 0
converges normally on any disc of radius strictly smaller than R.
Let r < R. We have known since Chap. II that the given series
fez) = I:anz n converges absolutely for Izl = r; since lanznl ::; lanrnl = Vn
for Izl ::; r, normal convergence follows and, incidentally, so again does the
continuity of the sum; but we knew that already!
The hypothesis r < R is essential. Consider for example the geometric
series I: zn which converges for Izl < 1 to the function s(z) = 1/(1 - z). We
shall show that the series does not converge norma.lly, nor even uniformly, on
the disc I z I < 1. If it did one could find an n such that
Is(z) - sn(z)1 ::; 1
on all the disc Izl < 1. The polynomial sn(z) = 1 + z + ... + zn being, in
absolute value, majorised by n + 1 for Izl < 1, one would then have Is(z)1 ::;
n + 2, i.e. 11 - zl ~ 1/(n + 2) for Izl < 1; impossible on a neighbourhood of
z = 1. Since furthermore Is(z) - sn(z)1 = Izln+l III - zl, it is clear that, for
n given, the left hand side increases indefinitely as Izl tends to 1, so is not
even bounded on Izl < 1.
If one restricts oneself temporarily to real values of z and if one remarks
that every compact, so closed, interval contained in X =] - 1, +1[ is in fact
contained in an interval [-q, +q] with q < 1, one has thus obtained a series
of continuous functions which, without converging uniformly on the open interval X, converges uniformly on every compact interval contained in X. A
very frequent phenomenon, valid for every power series and which nevertheless suffices to assure the continuity of the sum of the series. It is in fact
continuous on [-1, 1[ because it is equal to 1/(1 - z); the following example
will provide a better reason in a more interesting case.
Example 4. For x real consider the series
(8.7)
L(x) = x - x 2 /2 + x 3 /3 - ... = 2) _l)n+1xn In
n~l
whose sum is the function 10g(1 + x), as we shall see (nO 16, example 1). It
converges absolutely for Ixl < 1 since its general term is, in absolute value,
majorised by that of the corresponding geometric series. It converges for x = 1
also, since then it is the alternating harmonic series I:( _1)n+l In. Finally, it
diverges for x = -1 (harmonic series). Let us show that its sum is continuous
on the interval] - 1,1] where it is defined.