Rearranging Changes the Sum

Reordering a conditionally convergent series changes its sum — ln⁡2≈0.6931\ln 2 \approx 0.6931 becomes 32ln⁡2≈1.0397\frac{3}{2}\ln 2 \approx 1.0397. Twin of the Leibniz scene.

∑n=1∞(−1)n+1n=ln⁡2→(+,+,−)32ln⁡2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = \ln 2 \xrightarrow{(+,+,-)} \frac{3}{2}\ln 2

Examples & Exercises

Verify numerically that the (+,+,−)(+,+,-) rearrangement's partial sums tend to 32ln⁡2≈1.0397\frac{3}{2}\ln 2 \approx 1.0397. (Numerical check; rigorous proof omitted.)

32ln⁡2≈1.03972\frac{3}{2}\ln 2 \approx 1.03972
n=150n=150: the rearranged partial sums settle near 1.03971.0397.
Show proof
  1. Directly computing T150T_{150} gives ∣T150−32ln⁡2∣≈0.0050<0.01|T_{150} - \frac{3}{2}\ln 2| \approx 0.0050 < 0.01. (Numerical check)
  2. Raising nn to 300300 shrinks the error to about 0.00250.0025 — the direction of convergence is consistent, not a coincidence.
  3. The general formula is stated as known (idea level): a rearrangement repeating pp positive and qq negative terms sums to ln⁡2+12ln⁡pq\ln 2 + \frac{1}{2}\ln\frac{p}{q}. For p=2,q=1p = 2, q = 1 this gives 32ln⁡2\frac{3}{2}\ln 2. Its proof is not covered here.

Explain at idea level why rearranging an absolutely convergent series never changes its sum. (Idea level; proof omitted.)

If tails can be made arbitrarily small, order stops mattering.
Show proof
  1. If ∑an\sum a_n converges absolutely, then for every ε>0\varepsilon > 0 there is an NN with tail ∑n>N∣an∣<ε\sum_{n>N}|a_n| < \varepsilon. (Sketch only; rigorous proof omitted)
  2. A rearrangement's partial sums eventually contain the first NN terms of the original, so the two partial sums differ only by tail terms — less than 2ε2\varepsilon.
  3. Since ε\varepsilon is arbitrary, the two sums agree: when tails can be made arbitrarily small, reordering cannot change the sum.

Prove ∣Sn−ln⁡2∣≤1n+1\lvert S_n - \ln 2\rvert \leq \frac{1}{n+1} for the alternating harmonic partial sums.

∣Rn∣≤an+1=1n+1\lvert R_n\rvert \leq a_{n+1} = \frac{1}{n+1}
n=100n=100: the error is trapped inside 1/101≈0.00991/101 \approx 0.0099.
Show proof
  1. an=1/na_n = 1/n decreases to 00. Even partial sums increase, odd ones decrease: S2m+2−S2m=a2m+1−a2m+2>0S_{2m+2} - S_{2m} = a_{2m+1} - a_{2m+2} > 0 and S2m+3−S2m+1=−(a2m+2−a2m+3)<0S_{2m+3} - S_{2m+1} = -(a_{2m+2} - a_{2m+3}) < 0.
  2. The limit S=ln⁡2S = \ln 2 always lies between two consecutive partial sums SnS_n and Sn+1S_{n+1}, so ∣S−Sn∣≤∣Sn+1−Sn∣=an+1=1n+1|S - S_n| \le |S_{n+1} - S_n| = a_{n+1} = \frac{1}{n+1}.
  3. For n=100n = 100 this gives ∣S100−ln⁡2∣≤1101≈0.0099|S_{100} - \ln 2| \le \frac{1}{101} \approx 0.0099 — matching the caption's figure.

7. Power Series