Conditional convergence and rearrangements of infinite series

� Definition: A series which converges but does not converge absolutely is said to be condi-tionally convergent.

Examples: The alternating harmonic series,∑∞

n=1(−1)n/n is conditionally convergent.

For a power series with radius of convergence ρ, conditional convergence can happen (if atall) only at points lying on the circle of convergence. At all other points, the series eitherdiverges 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∑an is an infinite series and f : N→ N is a bijection, then the series

∞∑n=1

afn

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 eventerm, then taking the next two odd ones, then one even one, etc. So the individual terms ofthe 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, forinstance, 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 cer-tainly 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∑afn be a rearrangement of the absolutely convergent series∑

an. Then∞∑n=1

an = A ⇒∞∑n=1

afn = A.

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Proof: Let

sn =n∑k=1

ak and s̃n =n∑k=1

afn .

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 problemis with the first term.

So, suppose ε > 0 is given. Then there exists n1 such that N ≥ n1 ⇒ |sN − A| ≤ ε/2.Furthermore, since the original series converges absolutely, the sequence of remainders,

rn =∞∑

k=n+1

|ak|

converges to 0. This means that for some n2, N ≥ n2 ⇒ rN < ε/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 | = |∑D

afk|

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 | = |∑D

afk| ≤

∑D

|afk| ≤

∞∑N+1

|ak| < ε/2.

We’ve therefore shown that for any ε > 0, |s̃n−A| < ε for n sufficiently large, so s̃n → A �

Example: The alternating harmonic series

∞∑n=1

(−1)n−1 1

n= 1− 1

2+

1

3− · · ·

is easily seen to be the Taylor series expansion of log(1 + x) evaluated at x = 1. That is, itssum is log 2. As we know, this series is conditionally convergent. If we try the rearrangementabove (two odd terms followed by an even term), we get the series

1 +1

3− 1

2+

1

5+

1

7− 1

4+ · · · .

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We claim this series converges to 32

log 2. To see this, note that

log 2 = 1− 1

2+

1

3− 1

4+

1

5− 1

6+

1

7+ · · ·

1

2log 2 = 0 +

1

2+ 0− 1

4+ 0 +

1

6+ 0− 1

8· · ·,

where we’ve inserted some zeros which don’t affect the sum. Adding the two series term byterm (which is legal, since they both converge), we get

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2log 2 = 1 + 0 +

1

3− 1

2+

1

5+ 0 +

1

7· · ·,

which is the same as the rearrangement, once we discard the 0’s.

In what follows, we suppose that∑an is conditionally convergent, and we define

a+n =

{an : if an > 00 : otherwise

}, and a−n =

{an : if an < 00 : otherwise

},

and

s+n =

n∑k=1

a+k , s−n =

n∑k=1

a−k .

♣ Exercise: Both∑a+n and

∑a−n contain infinitely many nonzero terms.

♣ Exercise: Neither of the series∑a+n or

∑a−n converges. (Hint:

∑an is not absolutely

convergent.)

♣ Exercise: The sequence of partial sums s+n is not bounded above; the sequence s−n is not

bounded below.

Theorem: Suppose the series∑an is conditionally convergent and choose

any r ∈ R. Then there exists a rearrangement of the series converging tor.

Proof: Since s+n is monotone increasing, there exists a least index n1 such that s+

n1−1 ≤r < s+

n1. Assuming all the 0 terms of a+

n and a−n to have been discarded, the first n1 termsof our rearrangement are a+

1 + a+2 + · · ·+ a+

n1. Now start adding terms from a−n , stopping at

m1, the least index such that the resulting sum is less than r:

(a+1 + a+

2 + · · ·+ a+n1

) + (a−1 + a−2 + · · ·+ am1) < r.

Again, we can do this because the series of negative terms is not bounded below. Forsimplicity, write

Q1 = a+1 + a+

2 + · · ·+ a+n1, and R1 = a−1 + a−2 + · · ·+ am1 .

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Now we return to the series of positive terms and add just enough so that we exceed r: thatis, we choose n2 to be the smallest index greater than n1 such that if Q2 = a+

n1+1 + · · ·+ a+n2

,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 sumis less than r: R2 = a−m1+1 + · · ·+ a−m2

, and

Q1 +R1 +Q2 +R2 < r.

And we continue in this way. Our rearrangement is the series Q1 +R1 +Q2 +R2 + · · ·. Andwe claim it sums to r:

♣ Exercise: Both sequences {a+n } and {a−n } 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 ofthe rearrangement to this point is

a+1 + · · ·+ a+

nk+1−1 + ank+1,

where the addition of the final term makes the partial sum > r. Let Q̂k+1 = Qk+1− a+nk+1−1.

Then we haveQ1 +R1 + · · ·+ Q̂k+1 ≤ r < Q1 +R1 + · · ·+Qk+1.

ThenQ1 + · · ·+Qk+1 − r ≤ (Q1 + · · ·Qk+1)− (Q1 + · · ·+ Q̂k+1) = a+

nk+1.

Now suppose ε > 0 is given. (We had to get here sometime!). There exists N ∈ N such thatn,m ≥ N ⇒ a+

n , a−m ∈ Bε(0). So, in the case at hand, choosing k so that nk+1 ≥ N implies

|Q1 +R1 + · · ·+Qk+1 − r| < ε.

♣ Exercise: Finish the proof for EXTRA CREDIT. It’s a little tricky, since you have toshow that the sequence of partial sums of the rearranged series converges, not just that thesubsequence given by the Qn’s and Rn’s converges.

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