Conditional convergence and rearrangements of infinite series
Definition: A series which converges but does not converge absolutely is said to be conditionally convergent.
Examples: The alternating harmonic series,
P∞
n
n=1 (−1) /n
is conditionally convergent.
For a power series with radius of convergence ρ, conditional convergence can happen (if at
all) only at points lying on the circle of convergence. At all other points, the series either
diverges or is absolutely convergent.
So, in some sense, conditional convergence is rare; but as is often the case in mathematics,
these rare cases are interesting.
Definition: If
P
an is an infinite series and f : N → N is a bijection, then the series
∞
X
af n
n=1
is said to be a rearrangement of the original series.
Example: We can rearrange a series by taking first two odd terms, then adding one even
term, then taking the next two odd ones, then one even one, etc. So the individual terms of
the series would be
a1 , a3 , a2 , a5 , a7 , a4 , a9 , a11 , a6 , . . .
(This goes on forever, since we never run out of either even or odd terms.)
♣ Exercise: Find f (n) for this example.
Example: Something like
a1 , a3 , a5 , a7 , . . . , a2n+1 , . . . , a2 , a4 , a6 , a8 . . .
where all the odd numbers precede all the even ones, is not a rearrangement since, for
instance, there’s no n ∈ N satisfying f (n) = 2.
Clearly, a rearrangement changes the values of the partial sums of the series, so it is certainly possible that a rearrangement converges to something other than the original limit.
Somewhat surprisingly, this doesn’t happen if the series converges absolutely.
Theorem:
Let
P
an . Then
P
afn be a rearrangement of the absolutely convergent series
∞
X
an = A ⇒
n=1
∞
X
n=1
1
afn = A.
Proof: Let
sn =
n
X
ak and s̃n =
n
X
k=1
af n .
k=1
Then for any n, N ∈ N,
|s̃n − A| = |s̃n − sN + sN − A| ≤ |s̃n − sN | + |sN − A|.
Since sn → A, we can make the second term on the right as small as we want; the problem
is with the first term.
So, suppose > 0 is given. Then there exists n1 such that N ≥ n1 ⇒ |sN − A| ≤ /2.
Furthermore, since the original series converges absolutely, the sequence of remainders,
∞
X
rn =
|ak |
k=n+1
converges to 0. This means that for some n2 , N ≥ n2 ⇒ rN < /2 as well. In what follows,
we take and fix N ≥ n1 , n2 .
Since f is a bijection, for some k1 , k2 , . . . , kN , fk1 = 1, fk2 = 2, . . . fkN = N . Take K =
max{k1 , . . . , kN }. Then we have
{1, 2, 3, . . . , N } ⊆ {fk1 , fk2 , . . . , fkN }.
Now for any n ≥ K, we have
|s̃n − sN | = |
X
af k |
D
where D is the finite set consisting of the subscripts which do not match up with any of
{1, 2, . . . , N }.
Again using the absolute convergence, we have
|s̃n − sN | = |
X
af k | ≤
D
X
D
|afk | ≤
∞
X
|ak | < /2.
N +1
We’ve therefore shown that for any > 0, |s̃n − A| < for n sufficiently large, so s̃n → A Example: The alternating harmonic series
∞
X
1 1
1
(−1)n−1 = 1 − + − · · ·
n
2 3
n=1
is easily seen to be the Taylor series expansion of log(1 + x) evaluated at x = 1. That is, its
sum is log 2. As we know, this series is conditionally convergent. If we try the rearrangement
above (two odd terms followed by an even term), we get the series
1+
1 1 1 1 1
− + + − + ···.
3 2 5 7 4
2
We claim this series converges to
3
2
log 2. To see this, note that
log 2 = 1 −
1 1 1 1 1 1
+ − + − + + ···
2 3 4 5 6 7
1
1
1
1
1
log 2 = 0 + + 0 − + 0 + + 0 − · · ·,
2
2
4
6
8
where we’ve inserted some zeros which don’t affect the sum. Adding the two series term by
term (which is legal, since they both converge), we get
3
1 1 1
1
log 2 = 1 + 0 + − + + 0 + · · ·,
2
3 2 5
7
which is the same as the rearrangement, once we discard the 0’s.
P
In what follows, we suppose that
an is conditionally convergent, and we define
an : if an > 0
an : if an < 0
+
−
an =
, and an =
,
0 : otherwise
0 : otherwise
and
s+
n
=
n
X
s−
n
a+
k,
k=1
♣ Exercise: Both
P
a+
n and
P
=
n
X
a−
k.
k=1
a−
n contain infinitely many nonzero terms.
♣ Exercise: Neither of the series
convergent.)
P
a+
n or
P
a−
n converges. (Hint:
P
an is not absolutely
−
♣ Exercise: The sequence of partial sums s+
n is not bounded above; the sequence sn is not
bounded below.
P
Theorem: Suppose the series
an is conditionally convergent and choose
any r ∈ R. Then there exists a rearrangement of the series converging to
r.
+
Proof: Since s+
n is monotone increasing, there exists a least index n1 such that sn1 −1 ≤
+
−
r < s+
n1 . Assuming all the 0 terms of an and an to have been discarded, the first n1 terms
+
+
−
of our rearrangement are a+
1 + a2 + · · · + an1 . Now start adding terms from an , stopping at
m1 , the least index such that the resulting sum is less than r:
+
+
−
−
(a+
1 + a2 + · · · + an1 ) + (a1 + a2 + · · · + am1 ) < r.
Again, we can do this because the series of negative terms is not bounded below. For
simplicity, write
+
+
−
−
Q1 = a+
1 + a2 + · · · + an1 , and R1 = a1 + a2 + · · · + am1 .
3
Now we return to the series of positive terms and add just enough so that we exceed r: that
+
is, we choose n2 to be the smallest index greater than n1 such that if Q2 = a+
n1 +1 + · · · + an2 ,
we have
Q1 + R1 + Q2 > r.
And now we pick the index m2 to be the least index greater than m1 so that our new sum
−
is less than r: R2 = a−
m1 +1 + · · · + am2 , and
Q1 + R1 + Q2 + R2 < r.
And we continue in this way. Our rearrangement is the series Q1 + R1 + Q2 + R2 + · · ·. And
we claim it sums to r:
−
♣ Exercise: Both sequences {a+
n } and {an } converge to 0.
♣ Exercise: The process of adding more and more Qn and Rn never stops.
Suppose we’re at the point where we’ve just appended the term Qk+1 . The partial sum of
the rearrangement to this point is
+
a+
1 + · · · + ank+1 −1 + ank+1 ,
bk+1 = Qk+1 − a+
where the addition of the final term makes the partial sum > r. Let Q
nk+1 −1 .
Then we have
bk+1 ≤ r < Q1 + R1 + · · · + Qk+1 .
Q1 + R1 + · · · + Q
Then
bk+1 ) = a+
Q1 + · · · + Qk+1 − r ≤ (Q1 + · · · Qk+1 ) − (Q1 + · · · + Q
nk+1 .
Now suppose > 0 is given. (We had to get here sometime!). There exists N ∈ N such that
−
n, m ≥ N ⇒ a+
n , am ∈ B (0). So, in the case at hand, choosing k so that nk+1 ≥ N implies
|Q1 + R1 + · · · + Qk+1 − r| < .
♣ Exercise: Finish the proof for EXTRA CREDIT. It’s a little tricky, since you have to
show that the sequence of partial sums of the rearranged series converges, not just that the
subsequence given by the Qn ’s and Rn ’s converges.
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