Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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Every rearrangement of k0(1/2)k\sum_{k \ge 0} (-1/2)^{k} converges to 2/32/3

Example

Let r:=1/2r := -1/2 and consider k0rk\sum_{k \ge 0} r^{k}, with rkr^k the integer power (Integer powers ama^m), so that the first term is r0=1r^0 = 1. Then:

k=0(12)k  =  11(1/2)  =  23,\sum_{k=0}^{\infty} \Bigl(-\tfrac12\Bigr)^{k} \;=\; \frac{1}{1 - (-1/2)} \;=\; \frac{2}{3},

the series converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), and every rearrangement of it along a bijection of N\mathbb{N} (Rearrangement of a series along a bijection of N\mathbb{N}, and unconditional convergence) converges, again to 2/32/3.

This is the contrast case for the whole page. The alternating harmonic series (j0(1)j/(j+1)\sum_{j \ge 0} (-1)^{j}/(j+1) converges conditionally, with sum strictly between 1/21/2 and 11) has terms with the same alternating sign pattern, tending to 00 just as these do, and can be rearranged to any real whatever; this series cannot be rearranged to anything but 2/32/3. The difference is absolute convergence and nothing else, by For a series of real numbers, unconditional convergence and absolute convergence are the same property.

Facts & Assumptions

Given: r=1/2r = -1/2 and the sequence ak:=rka_k := r^{k} (Integer powers ama^m).

[L1]

Geometric series: for x<1|x| < 1 the series xk\sum x^k converges with sum 1/(1x)1/(1-x), the series starting at k=0k = 0 with first term x0=1x^0 = 1 (For r<1|r| < 1, k0rk=1/(1r)\sum_{k \ge 0} r^k = 1/(1-r), and for r1|r| \ge 1 the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Absolute value: xy=xy|xy| = |x|\,|y|, 1=1|1| = 1, and 1/2=1/2|-1/2| = 1/2 (Basic properties of the absolute value).

[L3]

Powers: x0=1x^0 = 1 and xn+1=xnxx^{n+1} = x^n x (Integer powers ama^m, Laws of integer exponents).

[L4]

The principle of induction on N\mathbb{N} (The principle of mathematical induction).

Verification

technique · direct
1.1

An induction gives rk=rk=(1/2)k|r^{k}| = |r|^{k} = (1/2)^{k} for every kk: at k=0k = 0 both sides are 11, and rk+1=rkr=rkr=(1/2)k(1/2)|r^{k+1}| = |r^{k} r| = |r^{k}|\,|r| = (1/2)^{k}(1/2).

L2L3L4
1.2

Since r=1/2<1|r| = 1/2 < 1, the series krk\sum_k r^{k} converges with sum 1/(1r)=1/(3/2)=2/31/(1-r) = 1/(3/2) = 2/3.

L1L2algebra
2.1

Since 1/2=1/2<1|1/2| = 1/2 < 1, the series krk=k(1/2)k\sum_k |r^{k}| = \sum_k (1/2)^{k} converges, with sum 1/(11/2)=21/(1 - 1/2) = 2; so krk\sum_k r^{k} converges absolutely.

step 1.1L1L2
3.1

By Dirichlet's rearrangement theorem, for every bijection σ\sigma of N\mathbb{N} the series krσ(k)\sum_k r^{\sigma(k)} converges, with the same sum 2/32/3.

step 2.1L5
4.1

So the series converges absolutely with sum 2/32/3, and every rearrangement of it converges to 2/32/3.

step 1.2step 2.1step 3.1

Remarks

Depends on

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