Riemann series rearrangement theorem

Statement

Consider a series:

$\sum _{k=1}^{\mathrm{\infty }}{a}_k$

(Note: the starting point of the summation does not matter for the theorem).

Suppose the series is a conditionally convergent series: it is a convergent series but not an absolutely convergent series, i.e., the series $\sum _{k=1}^{\mathrm{\infty }}\left|a_k\right|$ does not converge.

Then, the following are true:

1. $\underset{k\to \mathrm{\infty }}{lim}{a}_k=0$
2. The sub-series comprising only those $a_k$s that are positive diverges.
3. The sub-series comprising only those $a_k$s that are negative diverges.
4. Given any two elements $L\le U$ in $[-\mathrm{\infty },\mathrm{\infty }]$ (i.e., they could be real numbers, or $\pm \mathrm{\infty }$) there exists a rearrangement of the $a_k$s such that the limit inferior of the partial sums is $L$ and the limit superior of the partial sums is $U$. In particular, since we are allowed to set $L=U$, we can obtain a rearrangement that converges to any desired sum.

Related facts

• Levy-Steinitz theorem is a generalization to series of vectors in ${\mathbb{R}}^n$. The claim is that the set of possible sums of rearrangements of any series of vectors that are finite vectors, if non-empty, is an affine subspace of ${\mathbb{R}}^n$.

Proof

Proof of (1), (2), and (3)

(1) is true on account of convergence.

(2) and (3): We can show that if any one of the sums is finite, so is the other one, and both being finite would force absolute convergence.

Proof of (4)

We outline a constructive procedure to create the series. For simplicity, we will assume that none of the $a_k$s are 0. The proof can be modified somewhat to include the case of some of the $a_k$s being zero.

Case of finite values of limit superior and limit inferior

Setup:

• Arrange all the positive values among the $a_k$s in decreasing order. We will always pick from the list in the order arranged, i.e., we will always pick the largest positive value currently in the list when asked to pick from the list. The fact that we can list all the positive $a_k$s in decreasing order relies on $\underset{k\to \mathrm{\infty }}{lim}{a}_k=0$.
• Arrange all the negative values among the $a_k$s in decreasing order of magnitude (hence, increasing order). We will always pick from the list in the order arranged, i.e., we will always pick the largest magnitude negative value currently in the list when asked to pick from the list. The fact that we can list all the positive $a_k$s in increasing order relies on $\underset{k\to \mathrm{\infty }}{lim}{a}_k=0$.

Rearrangement creation:

• Begin by picking positive values (starting with the largest) and keep doing so until the partial sum so far is greater than $U$. Note that we always reach such a point after picking finitely many positive values because the sum of all positive terms of the series is $\mathrm{\infty }$.
• Now, start picking negative values (starting with the largest magnitude) till the overall partial sum (including the positive and negative values picked so far) is less than $L$. Again, this is possible because the negative value terms add up to $-\mathrm{\infty }$.
• Now, resume picking positive values till the overall partial sum is greater than or equal to $U$, then switch back to picking negative values, and so on.

Checking conditions:

• Every term gets picked exactly once: We note that, because both the positive and negative term sums diverge, it is the case that we cycle back and forth infinitely many times. Each time we cycle, we pick at least one new term from each side, so it's clear that every term gets picked. The picking procedure makes sure every time is picked exactly once.
• The limit superior is $U$: We know that the partial sums become greater than or equal to $U$ infinitely often, so the limit superior is at least $U$. The extent to which the partial sum can overshoot $U$ each time is bounded by the magnitude of the terms, which we have assumed approaches zero. This shows that the limit superior is exactly $U$.
• The limit inferior is $L$: Similar reasoning as with the limit superior.

Case where we allow infinity for one or both the values

• If the limit superior is $\mathrm{\infty }$ and the limit inferior is finite: In this case, we construct an increasing sequence of finite numbers $U_1,U_2,\dots ,$ that approaches $\mathrm{\infty }$ first (we can construct this sequence before looking at the $a_k$s). Now, we modify the construction of the rearrangement series as follows: instead of trying to cross $U$ every time with the positive terms, we try to cross $U_i$ with the positive terms at the $i^{th}$ stage.
• If the limit superior and limit inferior are both $\mathrm{\infty }$: We construct increasing sequences to $\mathrm{\infty }$ in place of both $U$ and $L$.
• If the limit inferior is $-\mathrm{\infty }$ and the limit superior is finite: In this case, we construct a decreasing sequence of finite numbers $L_1,L_2,\dots$ that approaches $-\mathrm{\infty }$ first (we can construct this sequence without looking at the $a_k$s). Now, we modify the construction of the rearrangement series as follows: instead of trying to cross $L$ every time with the negative terms, we try to cross $L_i$ with the negative terms at the $i^{th}$ stage.
• If the limit superior and limit inferior are both $-\mathrm{\infty }$: We construct decreasing sequences to $-\mathrm{\infty }$ in place of both $U$ and $L$.
• If the limit superior is $\mathrm{\infty }$ and the limit inferior is $-\mathrm{\infty }$: We construct an increasing sequence to $\mathrm{\infty }$ in place of $U$ and a decreasing sequence to $-\mathrm{\infty }$ in place of $L$.