Rearrangement of Series; Riemann’s Theorem

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2017)
• Priority tier: T4
• Marks (count): 20 (1)
• Average solve time: ~25 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

This is a T4 atom — it has appeared only once in 13 years (2017, 20 marks). The topic draws a sharp conceptual line between absolute and conditional convergence: absolute convergence is robust under permutation of terms, while conditional convergence is fragile in the most dramatic way possible. When UPSC does ask, they want a clean proof of Riemann’s rearrangement theorem or the absolute-convergence rearrangement result, so knowing the argument cold is worth the investment.

Minimum Theory

Setup. A rearrangement of $\sum_{n=1}^{\infty} a_n$ is a series $\sum_{n=1}^{\infty} a_{\sigma(n)}$, where $\sigma:\mathbb{N}\to\mathbb{N}$ is a bijection.

Theorem A (Absolute convergence is rearrangement-invariant). If $\sum a_n$ converges absolutely, then every rearrangement converges to the same sum $S$.

Proof. Let $S = \sum_{n=1}^{\infty} a_n$ and let $\sum a_{\sigma(n)}$ be any rearrangement. Let $\varepsilon > 0$. Since $\sum |a_n|$ converges, choose $N$ such that $\sum_{n=N+1}^{\infty} |a_n| < \varepsilon$. Choose $M$ large enough so that $\{1, 2, \ldots, N\} \subseteq \{\sigma(1), \sigma(2), \ldots, \sigma(M)\}$. Then for all $m \geq M$,

$\left|\sum_{k=1}^{m} a_{\sigma(k)} - S\right| = \left|\sum_{k=1}^{m} a_{\sigma(k)} - \sum_{n=1}^{\infty} a_n\right| \leq \sum_{n \notin \{\sigma(1),\ldots,\sigma(m)\}} |a_n| \leq \sum_{n=N+1}^{\infty} |a_n| < \varepsilon.$

Since $\varepsilon$ was arbitrary, $\sum a_{\sigma(n)} = S$. $\blacksquare$

Theorem B (Riemann’s Rearrangement Theorem). If $\sum a_n$ converges conditionally (i.e., $\sum a_n$ converges but $\sum |a_n| = \infty$), then for any $S \in \mathbb{R} \cup \{+\infty, -\infty\}$ there exists a rearrangement of $\sum a_n$ that converges to $S$.

Key Lemma. Define the positive and negative parts:

$p_n = \frac{a_n + |a_n|}{2} = \max(a_n, 0), \qquad q_n = \frac{|a_n| - a_n}{2} = \max(-a_n, 0).$

Then $a_n = p_n - q_n$ and $|a_n| = p_n + q_n$. Since $\sum a_n$ converges but $\sum |a_n|$ diverges, both $\sum p_n = +\infty$ and $\sum q_n = +\infty$. (If either were finite, the other would too, forcing $\sum |a_n|$ to converge — contradiction.) Also $p_n \to 0$ and $q_n \to 0$ since $a_n \to 0$.

Proof of Theorem B (for $S \in \mathbb{R}$). Enumerate the positive terms as $P_1, P_2, \ldots$ (in order of original index) and the negative terms as $-Q_1, -Q_2, \ldots$ (each $Q_i > 0$). Construct the rearrangement as follows:

• Step 1. Take the fewest positive terms $P_1 + P_2 + \cdots + P_{m_1}$ such that the partial sum first exceeds $S$.
• Step 2. Then append the fewest negative terms $-Q_1 - Q_2 - \cdots - Q_{n_1}$ such that the partial sum first falls below $S$.
• Step k (odd). Append more positive terms until the sum first exceeds $S$.
• Step k (even). Append more negative terms until the sum first falls below $S$.

This process never stalls because $\sum P_n = \sum Q_n = +\infty$. Let $A_k$ denote the partial sum at the end of step $k$. At an “overshoot” step, $A_k > S$ and $A_k \leq S + P_{m_k}$; at an “undershoot” step, $A_k < S$ and $A_k \geq S - Q_{n_k}$. Since $P_n \to 0$ and $Q_n \to 0$, the oscillation $|A_k - S|$ is bounded by the last term added, which $\to 0$. Therefore $A_k \to S$, and every partial sum lies between two consecutive milestone partial sums whose distance to $S$ tends to zero. Hence the full sequence of partial sums converges to $S$. $\blacksquare$

Corollary. A rearrangement can also be constructed to diverge to $+\infty$ or $-\infty$ by taking ever more positive or negative terms without switching.

Question Archetypes

prove-absolute-rearrangement (1 question(s); 2017)

Recognition Cues

• The question mentions “absolutely convergent” and “rearrangement.”
• May ask you to prove the invariance of the sum, or to distinguish from the conditional case.

Solution Template

1. State the theorem precisely (bijection $\sigma$, same limit $S$).
2. Use $\varepsilon$-argument: cover $\{1,\ldots,N\}$ inside $\{\sigma(1),\ldots,\sigma(M)\}$ for large $M$.
3. Bound tail by $\sum_{n > N} |a_n| < \varepsilon$.
4. Conclude $\sum a_{\sigma(n)} = S$. $\blacksquare$

Worked Example

2017 Paper 2, 2017-P2-Q1b (20 marks)

State and prove Riemann’s rearrangement theorem for real series.

Claim. If $\sum a_n$ converges conditionally, then for any $S \in \mathbb{R}$ there is a rearrangement converging to $S$.

Proof.

Step 1 — Auxiliary divergence. Since $\sum a_n$ converges, $a_n \to 0$, so $p_n = \max(a_n,0) \to 0$ and $q_n = \max(-a_n,0) \to 0$. Since $\sum a_n$ converges conditionally, $\sum |a_n| = \infty$, hence $\sum p_n = \infty$ and $\sum q_n = \infty$ (as argued in the lemma above).

Step 2 — Construction. List the positive summands as $P_1 \geq 0, P_2 \geq 0, \ldots$ and the negative summands by their magnitudes $Q_1, Q_2, \ldots > 0$, each sequence retaining original order. Define the rearrangement inductively:

• Choose minimal $m_1$ with $P_1 + \cdots + P_{m_1} > S$.
• Choose minimal $n_1$ with $P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{n_1} < S$.
• Alternate: add positive terms until first exceeding $S$, then negative terms until first falling below $S$.

Such indices always exist because $\sum P_n = \sum Q_n = \infty$.

Step 3 — Convergence. Let $s_k$ be the partial sum after stage $k$. After an overshoot stage, $s_k \in (S, S + P_{m_k}]$; after an undershoot stage, $s_k \in [S - Q_{n_k}, S)$. Since $P_n \to 0$ and $Q_n \to 0$, we have $|s_k - S| \to 0$.

Every partial sum $s$ of the rearrangement lies between two consecutive milestone sums $s_k$ and $s_{k+1}$, both within $\max(P_{m_{k+1}}, Q_{n_{k+1}})$ of $S$, which tends to $0$. Hence $s_k \to S$.

$\boxed{\text{Every conditionally convergent series can be rearranged to any prescribed sum.}}\;\blacksquare$

Common Traps

• Forgetting to prove $\sum p_n = \sum q_n = \infty$; this is the key lemma and must be stated.
• Confusing “the process never stalls” with “the process terminates”; emphasise that because the partial sums of $P_n$ and $Q_n$ both diverge, each stage completes in finitely many terms.
• Omitting the argument that all intermediate partial sums (not just milestone ones) also converge to $S$.

Marks-Aware Writing

At 20 marks this is a full Section B proof question. Allocate roughly: (a) state the theorem precisely — 3 marks; (b) prove the auxiliary lemma ($\sum p_n = \sum q_n = \infty$) — 5 marks; (c) construct the rearrangement — 6 marks; (d) prove convergence of the rearrangement — 6 marks. Write in complete sentences with logical connectives; do not list steps without justification.

Practice Set

Only one historical question on this atom (shown above).