§2. Absolutely convergent series
143
already studied in nO 12 again. A possible way of partitioning 1= 'Z} - {O},
where 0 denotes the pair (0,0), consists of grouping all the terms for which
m has a given value, Le. all the points of the lattice Z2 situated on the same
vertical in the plane. Unconditional convergence then reduces to the relation
(18.12)
L L 1/(m 2 + n 2 )k/2 < +00
mEZ nEZ
since all the terms are positive; it goes without saying that one excludes the
pair (0,0). Convergence as a series in n is clear for k > 1 since, for m given,
one manifestly has (m 2 + n 2 )k/2 :::::: Inlk; but the sum, which we do not know,
depends on m in a way too little obvious for (12) to be usable here. One might
also interchange the roles of m and n, Le. reverse the order of summations in
(12), but one would not advance further for all that ...
A more natural partition consists of grouping all the pairs (m, n) for which
m 2 +n 2 has a given value p ~ 1. There are only finitely many such, since then
Iml,lnl ::; p. Let N(p) this number; the corresponding partial sum is clearly
equal to N(p)/pk/2, so instead of verifying (12) one might instead attempt
to verify that
(18.13)
L N(p)/pk/2 < +00,
which brings us back to the case of a simple series; for this method to be
usable one would need to know, if not the exact value, at least an order of
magnitude of N(p) for p large. This is unfortunately not obvious 43 • Instead
of considering the pairs (m, n) lying on a given circumference with centre 0,
one could also group those for which Iml + Inl has a given value p, whence
another translation of unconditional convergence:
00
(18.14)
L
p=l
L u(m,n) = lim L u(m,n) < +00.
p-+oo
Iml+lnl=p
Iml+lnl:5p
One might imagine all sorts of other more or less bizarre groupings of terms,
for example grouping together the terms situated on the same equilateral
hyperbola m 2 - n 2 = p, but such a grouping would hardly be very well
adapted to the situation.
As we saw in nO 12 by using ad hoc arguments, version (14) is the best
adapted to the problem. The terms for which Iml + Inl has a given value pall
have the same order of magnitude, namely l/pk, and there are 4p of them.
One is thus led to the Riemann series E l/pk-l, whence the condition k > 2
for convergence.
43 The calculation of N (p) is a very interesting problem in number theory. The first
remark to make is that N(p) is frequently zero. In fact, N(p) = 4(n~-n;) where
n~ (resp. n;) is the number of divisors of p of the form 4k + 1 (resp. 4k + 3).
143
already studied in nO 12 again. A possible way of partitioning 1= 'Z} - {O},
where 0 denotes the pair (0,0), consists of grouping all the terms for which
m has a given value, Le. all the points of the lattice Z2 situated on the same
vertical in the plane. Unconditional convergence then reduces to the relation
(18.12)
L L 1/(m 2 + n 2 )k/2 < +00
mEZ nEZ
since all the terms are positive; it goes without saying that one excludes the
pair (0,0). Convergence as a series in n is clear for k > 1 since, for m given,
one manifestly has (m 2 + n 2 )k/2 :::::: Inlk; but the sum, which we do not know,
depends on m in a way too little obvious for (12) to be usable here. One might
also interchange the roles of m and n, Le. reverse the order of summations in
(12), but one would not advance further for all that ...
A more natural partition consists of grouping all the pairs (m, n) for which
m 2 +n 2 has a given value p ~ 1. There are only finitely many such, since then
Iml,lnl ::; p. Let N(p) this number; the corresponding partial sum is clearly
equal to N(p)/pk/2, so instead of verifying (12) one might instead attempt
to verify that
(18.13)
L N(p)/pk/2 < +00,
which brings us back to the case of a simple series; for this method to be
usable one would need to know, if not the exact value, at least an order of
magnitude of N(p) for p large. This is unfortunately not obvious 43 • Instead
of considering the pairs (m, n) lying on a given circumference with centre 0,
one could also group those for which Iml + Inl has a given value p, whence
another translation of unconditional convergence:
00
(18.14)
L
p=l
L u(m,n) = lim L u(m,n) < +00.
p-+oo
Iml+lnl=p
Iml+lnl:5p
One might imagine all sorts of other more or less bizarre groupings of terms,
for example grouping together the terms situated on the same equilateral
hyperbola m 2 - n 2 = p, but such a grouping would hardly be very well
adapted to the situation.
As we saw in nO 12 by using ad hoc arguments, version (14) is the best
adapted to the problem. The terms for which Iml + Inl has a given value pall
have the same order of magnitude, namely l/pk, and there are 4p of them.
One is thus led to the Riemann series E l/pk-l, whence the condition k > 2
for convergence.
43 The calculation of N (p) is a very interesting problem in number theory. The first
remark to make is that N(p) is frequently zero. In fact, N(p) = 4(n~-n;) where
n~ (resp. n;) is the number of divisors of p of the form 4k + 1 (resp. 4k + 3).