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Taking two positive terms for each negative one rearranges the alternating harmonic series to $3/2$ times its sum, by the identity $T_{3n} = S_{4n} + \tfrac12 S_{2n}$

Example

Let $a_j := (-1)^j/\iota(j+1)$ be the terms of the alternating harmonic series, whose sum $S$ satisfies $1/2 < S < 1$ ( $\sum_{j \ge 0} (-1)^{j}/(j+1)$ converges conditionally, with sum strictly between $1/2$ and $1$ ), and let $S_N := \sum_{j<N} a_j$ be its partial sums.

Rearrange it by taking two positive terms for each negative one:

$1 + \tfrac13 - \tfrac12 \;+\; \tfrac15 + \tfrac17 - \tfrac14 \;+\; \tfrac19 + \tfrac1{11} - \tfrac16 \;+\; \cdots$

Formally, define $\sigma : \mathbb{N} \to \mathbb{N}$ by

$\sigma(3m) = 4m, \qquad \sigma(3m+1) = 4m+2, \qquad \sigma(3m+2) = 2m+1 \qquad (m \in \mathbb{N}),$

which is a bijection (Injection, surjection, bijection), so that $\sum_k a_{\sigma(k)}$ is a rearrangement of the alternating harmonic series (Rearrangement of a series along a bijection of $\mathbb{N}$ , and unconditional convergence). Writing $T_n := \sum_{k<n} a_{\sigma(k)}$ , the identity

$T_{3n} \;=\; S_{4n} \;+\; \tfrac12\,S_{2n} \qquad (n \in \mathbb{N})$

holds, and consequently

$\sum_{k=0}^{\infty} a_{\sigma(k)} \;=\; \tfrac32\,S .$

The value is stated relative to $S$ , and deliberately so. Texts that already have the logarithm state this example as a multiple of $\log 2$ ; that expression is not available at this point in the reading order, and the identity above needs none (Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for). Since $1/2 < S < 1$ , the rearranged sum lies strictly between $3/4$ and $3/2$ , and in particular differs from $S$ : a concrete instance of The Riemann series theorem: a conditionally convergent real series has, for every $c \in \mathbb{R}$ , a rearrangement with sum $c$ , and rearrangements diverging to $+\infty$ , to $-\infty$ , and oscillating with any prescribed $\liminf \le \limsup$ in $\overline{\mathbb{R}}$ .

Facts & Assumptions

Given: The alternating sequence $(\varepsilon_j)$ with index maps $e_j = 2j$ and $o_j = 2j+1$ ; the terms $a_j = \varepsilon_j/\iota(j+1)$ ; the partial sums $S_N = \sum_{j<N}a_j$ of the alternating harmonic series, with sum $S$ ; and $T_n = \sum_{k<n}a_{\sigma(k)}$ .

[L1]

The alternating sequence and its index maps: $e_0 = 0$ , $e_{j+1} = e_j+2$ , $o_0 = 1$ , $o_{j+1} = o_j+2$ ; $\mathbb{N}$ is the disjoint union of the ranges of $e$ and $o$ , each element occurring for exactly one index; $\varepsilon_{e_j} = 1$ and $\varepsilon_{o_j} = -1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ).

[L2]

The alternating harmonic series converges, with sum $S$ satisfying $1/2 < S < 1$ ( $\sum_{j \ge 0} (-1)^{j}/(j+1)$ converges conditionally, with sum strictly between $1/2$ and $1$ , The alternating series test: if $(b_k)$ is nonincreasing with $b_k \to 0$ then $\sum_{k} (-1)^{k} b_k$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_n$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

The canonical naturals are positive for $n \ge 1$ and strictly increasing; reciprocation reverses the order on the positives; and for every real $\varepsilon > 0$ there is $n \ge 1$ with $1/\iota(n) < \varepsilon$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L4]

The recursion theorem and the principle of induction (The recursion theorem, The principle of mathematical induction); every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L5]

Finite sums: $\sum_{k<0}x_k = 0$ , $\sum_{k<n+1}x_k = \sum_{k<n}x_k + x_n$ , additivity, scaling and splitting (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

A subsequence of a convergent sequence converges to the same limit (Subsequences inherit the limit).

[L7]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L8]

A rearrangement is the composite of the terms with a bijection of $\mathbb{N}$ (Rearrangement of a series along a bijection of $\mathbb{N}$ , and unconditional convergence, Injection, surjection, bijection).

Verification

technique · direct

1.1

An induction gives $e_j = 2j$ and $o_j = 2j+1$ for every $j$ , from $e_0 = 0$ , $o_0 = 1$ and the two recursions; so by [L1] every natural number is $2i$ for exactly one $i$ or $2i+1$ for exactly one $i$ .

L1L4

1.2

Every natural $n$ is $3m + r$ for exactly one pair $(m, r)$ with $r < 3$ : for existence, the set $\{\, m' : 3m' > n \,\}$ is nonempty, containing $n+1$ , so it has a least element $m_0$ , which is not $0$ since $3 \cdot 0 = 0 \le n$ ; put $m := m_0 - 1$ , so $3m \le n < 3m+3$ and $r := n - 3m$ satisfies $r < 3$ . For uniqueness, if $3m + r = 3m' + r'$ with $r, r' < 3$ and $m < m'$ , then $3m + r < 3m + 3 \le 3m' \le 3m' + r'$ , a contradiction; so $m = m'$ and then $r = r'$ .

1.3

The maps $n \mapsto 4n$ and $n \mapsto 2n$ are strictly increasing, so $(S_{4n})_n$ and $(S_{2n})_n$ are subsequences of $(S_N)$ and both converge to $S$ .

L2L6

2.1

Applying step 1.1 twice, every natural number is exactly one of $4i$ , $4i+2$ or $2i+1$ , for exactly one $i$ : an even number $2m$ is $4i$ when $m = 2i$ and $4i+2$ when $m = 2i+1$ , and these two cases are exclusive and exhaustive by step 1.1 applied to $m$ .

step 1.1

3.1

The map $\sigma$ is therefore a well-defined function on $\mathbb{N}$ , given on the unique representation $n = 3m+r$ by the three clauses of the statement; it may equally be produced by the recursion theorem applied to the state set $\mathbb{N} \times \{0,1,2\}$ with the cycle $(m,0) \mapsto (m,1) \mapsto (m,2) \mapsto (m+1,0)$ . It is a bijection: by step 1.2 the pairs $(m,r)$ with $r<3$ correspond exactly to the naturals $n = 3m+r$ , and by step 2.1 the three clauses send those pairs bijectively onto $\mathbb{N}$ .

step 2.1step 1.2L4L8

4.1

By [L1] and step 1.1, $\varepsilon_{4m} = \varepsilon_{4m+2} = 1$ and $\varepsilon_{2m+1} = -1$ , so $a_{\sigma(3m)} = 1/\iota(4m+1)$ , $a_{\sigma(3m+1)} = 1/\iota(4m+3)$ and $a_{\sigma(3m+2)} = -1/\iota(2m+2)$ .

step 1.1step 3.1L1

5.1

An induction on $n$ gives $T_{3n} = S_{4n} + \tfrac12 S_{2n}$ . At $n = 0$ all three sums are empty, hence $0$ . For the step, by step 4.1 and [L5], $T_{3(n+1)} - T_{3n} = 1/\iota(4n+1) + 1/\iota(4n+3) - 1/\iota(2n+2)$ , while $S_{4(n+1)} - S_{4n} = 1/\iota(4n+1) - 1/\iota(4n+2) + 1/\iota(4n+3) - 1/\iota(4n+4)$ and $\tfrac12\bigl(S_{2(n+1)} - S_{2n}\bigr) = \tfrac12\bigl(1/\iota(2n+1) - 1/\iota(2n+2)\bigr) = 1/\iota(4n+2) - 1/\iota(4n+4)$ ; adding the last two gives $1/\iota(4n+1) + 1/\iota(4n+3) - 2/\iota(4n+4)$ , and $2/\iota(4n+4) = 1/\iota(2n+2)$ .

step 4.1L3L4L5

5.2

For $3n \le m \le 3n+2$ the difference $T_m - T_{3n}$ is a sum of at most the two positive terms $1/\iota(4n+1)$ and $1/\iota(4n+3)$ , so $0 \le T_m - T_{3n} \le 2/\iota(4n+1)$ .

step 4.1L3L5

6.1

Hence $T_{3n} \to S + \tfrac12 S = \tfrac32 S$ by step 5.1 and the algebra of limits.

step 5.1step 1.3L7

7.1

Let $\varepsilon > 0$ be rational. By step 6.1 fix $N_1$ with $|T_{3n} - \tfrac32 S| < \varepsilon/2$ for $n \ge N_1$ , and by [L3] fix $N_2 \ge 1$ with $2/\iota(4N_2+1) < \varepsilon/2$ , which then holds with $n$ in place of $N_2$ for every $n \ge N_2$ ; put $N := \max\{N_1, N_2\}$ .

step 6.1L3choose

8.1

Let $m \ge 3N$ and write $m = 3n + r$ with $r < 3$ as in step 1.2; then $3N \le m < 3n+3$ , so $N < n+1$ and $n \ge N$ . Hence $|T_m - \tfrac32 S| \le |T_m - T_{3n}| + |T_{3n} - \tfrac32 S| < \varepsilon/2 + \varepsilon/2 = \varepsilon$ .

step 1.2step 5.2step 7.1

9.1

Therefore $T_m \to \tfrac32 S$ : the rearranged series $\sum_k a_{\sigma(k)}$ converges with sum $\tfrac32 S$ , and since $1/2 < S < 1$ that sum lies strictly between $3/4$ and $3/2$ , so in particular it is not $S$ .

step 3.1step 8.1L2∎

Remarks

Why the identity is the right thing to prove. It compares the rearranged partial sums with two subsequences of the original partial sums, and both subsequences converge to $S$ for free. No estimate of $S$ is needed anywhere, and no closed form for it; the whole computation is an exact identity between finite sums, checked at $n = 1$ as $1 + 1/3 - 1/2 = 5/6$ against $7/12 + 1/4 = 5/6$ .
The rearrangement is explicit, unlike the one produced by The Riemann series theorem: a conditionally convergent real series has, for every $c \in \mathbb{R}$ , a rearrangement with sum $c$ , and rearrangements diverging to $+\infty$ , to $-\infty$ , and oscillating with any prescribed $\liminf \le \limsup$ in $\overline{\mathbb{R}}$ . There the bijection is built by a greedy recursion depending on the whole series; here it is given by three formulas. The price is that its sum is whatever the identity says it is, rather than a prescribed target.
The three residual index classes are handled, not waved through. The identity constrains $T_m$ only at multiples of $3$ ; step 5.2 and step 8.1 close the gap, using that the two intervening terms are positive and tend to $0$ . An argument stopping at step 6.1 would have proved convergence of a subsequence only.

Depends on

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Sources

Riemann series theorem (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
N. Donaldson, Math 140A: Series (standard reference, not scraped)

Taking two positive terms for each negative one rearranges the alternating harmonic series to 3/23/23/2 times its sum, by the identity T3n=S4n+12S2nT_{3n} = S_{4n} + \tfrac12 S_{2n}T3n​=S4n​+21​S2n​