Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(2x)^n}n. $$
Explanation of Method
This is a power series in $x$ with coefficients $a_n=(-1)^{n-1}2^n/n$. The ratio test locates the radius of convergence at $|x|=\frac12$, and the two endpoints must be checked individually — one reduces to the alternating harmonic series (conditionally convergent), the other to the negative harmonic series (divergent). The series is in fact $\ln(1+2x)$.
- Find the radius of convergence via the ratio test.
- Test both endpoints $x=\pm\frac12$ individually.
- Identify the series as $\ln(1+2x)$ and summarize the convergence regions.
Step 1 · Radius of convergence
$\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{2n}{n+1}\to2$, so the series (in $x$) converges when $2|x|<1$, i.e. $|x|<\frac12$, and diverges for $|x|>\frac12$.
Step 2 · Test the endpoints
At $x=\frac12$: the series becomes $\sum(-1)^{n-1}/n$, the alternating harmonic series — convergent (Leibniz), but $\sum1/n$ diverges, so conditionally convergent. At $x=-\frac12$: the series becomes $\sum(-1)^{n-1}(-1)^n/n=\sum(-1)^{2n-1}/n=-\sum1/n$, which diverges.
Step 3 · Identify the series
Recall $\ln(1+y)=\sum_{n=1}^{\infty}(-1)^{n-1}y^n/n$ for $-1<y\le1$. With $y=2x$: the series equals $\ln(1+2x)$, valid exactly for $-\frac12<x\le\frac12$, matching the convergence analysis.
Summary Table
| Region | Behavior |
|---|---|
| $|x|<\frac12$ | Absolutely convergent |
| $x=\frac12$ | Conditionally convergent |
| $x=-\frac12$ or $|x|>\frac12$ | Diverges |
Additional Notes
- This series shares its underlying constant ($\ln2$) with Problem 154's double series, since setting $x=\frac12$ recovers exactly the alternating harmonic series representation of $\ln2$.
- The asymmetric endpoint behavior (convergent at $x=\frac12$, divergent at $x=-\frac12$) is characteristic of the logarithm's power series, which has a genuine singularity at $y=-1$ (i.e. $x=-\frac12$) but only a conditionally-convergent boundary point at $y=1$.
- This is a standard illustration that a power series' interval of convergence can be a half-open interval, with the endpoint behavior determined entirely by separate tests (Leibniz's vs. direct comparison) rather than by the ratio test itself, which is inconclusive exactly at the radius.
Result
The convergence regions are: