Given that sum from n=1 to infinity ((-1)^(n-1))/n=ln 2
Given that $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=\ln 2$, find the sum of the series $$ 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k}+\cdots $$ which is obtained from the given series by rearrangement of its terms.
Explanation of Method
This rearrangement groups the alternating harmonic series into blocks of three: one positive odd-reciprocal term followed by two negative even-reciprocal terms. Since the alternating harmonic series converges only conditionally, Riemann's rearrangement theorem guarantees such reorderings can change the sum — and the classical $(p,q)$-rearrangement formula (taking $p$ positive terms then $q$ negative terms in a repeating pattern) gives the exact new sum in terms of $\ln2$ and $\ln(p/q)$.
- Identify the $k$-th group $\frac1{2k-1}-\frac1{4k-2}-\frac1{4k}$ and simplify it.
- Recognize the pattern as a $(1,2)$-rearrangement of the alternating harmonic series.
- Apply the known rearrangement formula to compute the new sum.
Step 1 · Simplify the $k$-th group
$$ \frac1{2k-1}-\frac1{4k-2}-\frac1{4k} = \frac1{2k-1}-\frac1{2(2k-1)}-\frac1{4k} = \frac1{2(2k-1)}-\frac1{4k} = \frac1{4k(2k-1)}. $$
Step 2 · Recognize the rearrangement pattern
The rearranged series takes one positive term from the odd reciprocals, followed by two negative terms from the even reciprocals, repeated indefinitely — a $(p,q)=(1,2)$ rearrangement of the alternating harmonic series.
Step 3 · Apply the rearrangement formula
For a $(p,q)$-rearrangement of $\sum\frac{(-1)^{n-1}}n=\ln2$ (taking $p$ positive terms, then $q$ negative terms, repeating), the classical result gives a new sum of $\ln2+\frac12\ln\!\left(\frac pq\right)$. With $p=1,q=2$:
$$ \text{Sum} = \ln2 + \frac12\ln\!\left(\frac12\right) = \ln2 - \frac12\ln2 = \frac{\ln2}2. $$
Summary Table
| Quantity | Value |
|---|---|
| Original sum ($p=q=1$) | $\ln2$ |
| Rearrangement pattern | $(p,q)=(1,2)$ |
| New sum | $\dfrac{\ln2}2$ |
Additional Notes
- This is a direct, concrete manifestation of Riemann's rearrangement theorem: because the alternating harmonic series is only conditionally convergent (Problem 103, $\alpha=1$), reordering its terms — even by such a simple, regular pattern — genuinely changes the sum from $\ln2\approx0.693$ down to $\frac{\ln2}2\approx0.347$.
- The general $(p,q)$-rearrangement formula, $\ln2+\frac12\ln(p/q)$, shows that taking more negative terms per block than positive ($q>p$) always shrinks the sum below $\ln2$, while taking more positive terms per block ($p>q$) would increase it — the extreme case $q\to\infty$ (all-negative tail) would drive the sum to $-\infty$.
- By contrast, no such rearrangement effect is possible for an absolutely convergent series (like Problem 104's telescoping alternating series, which sums to exactly $-1$ regardless of any reordering) — this dichotomy is precisely what separates absolute from conditional convergence in practice.
Result
The rearranged series converges to $\dfrac{\ln2}2$, half the original sum: