26 Reminders concerning convergen ce
1.3.d Conditionally convergent series,
absolutely convergent series
DEFINITION 1.49 A series
P
a
n
with a
n
in a normed vector space is condi-
tionnally convergent if it is convergent but not absolutely convergent.
DEFINITION 1.50 We denote by S the group of permuta tions, that is, the
group of bijections from N to N, and we denote by S
n
the finite group of
permutations of the set {1, . . ., n}.
DEFINITION 1.51 A series
P
x
n
is commutatively convergent if it is con-
vergent with sum X , and for any permutat ion ϕ ∈ S, the rearranged series
P
x
ϕ(n)
converges to X .
Is a convergent series necessarily commutatively convergent? I n other words,
is it legitimate to change arbitrarily the order of the terms of a convergent series?
At first sight, it is very tempting to say “Yes ,” almost without thinking,
since permuting terms in a finite sum has no effect on th e result. The problem
is that we have here an infinite sum, not a finite sum in the algebraic sense.
So there is a limiting process involved, and we will see th at this brings a
very different picture: only absolutely convergent series will be commutatively
convergent. So, if a series is conditionally convergent (convergent but not
absolutely so), changing the order of the terms may alter the value of the sum
— or e ven turn it into a divergent series.
THEOREM 1.52 Let
P
a
n
be a conditionnally convergent series, with a
n
∈ R. Th en
for any ℓ ∈ R, there exists a permutation ϕ ∈ S such that the rearranged series
P
a
ϕ(n)
converges to ℓ.
In fact, for any a, b ∈ R, with a ¶ b, there exists a permuta tion ψ ∈ S such that
the set of limit points of the sequence of rearranged partial sums
P
n
k=0
a
ψ(k)
n∈N
is
the interval [a, b].
Proof. We assume that
P
a
n
is conditionnally convergent.
◮ First remark: there are infinitely many positive values and infinitely many negative values
of the terms a
n
of the series. Let α
n
denote the sequence of non-negative terms, in the
order they occur, and let β
n
denote the sequence of negative terms.
Here is an illustration:
u
n
: 1 3 2 −4 −1 2 −1 0 2 ···
α
1
α
2
α
3
β
1
β
2
α
4
β
3
α
5
α
6
···
◮ Second remark: both series
P
α
n
and
P
β
n
are divergent, their partial sums con-
verging, respectively, to +∞ and −∞. Indeed, if both series were to converge, the
series
P
a
n
would be absolutely convergent, and if only one were to converge, then
P
a
n
would be divergent (as follows from considering the sequence of partial sums).
◮ Third remark: both sequences (α
n
)
n∈N
and (β
n
)
n∈N
tend to 0 (since (a
n
)
n∈N
tends
to 0, as a consequence of the convergence of
P
a
n
).
Now consider ℓ ∈ R. Let S
n
denote the sequence of sums of values of α and β
which is constructed as follows. First, sum all consecutive values of α
n
until their sum