Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}n$.
Explanation of Method
This is the Taylor series of $\ln(1+x)$, with radius $1$. At $x=1$ it reduces to the alternating harmonic series (converges conditionally); at $x=-1$ every sign flips out and it reduces to the negative of the harmonic series (diverges) — the mirror image of Problem 225.
- Apply the ratio test to find $R$.
- Test $x=1$: the alternating harmonic series.
- Test $x=-1$: reduces to the (negated) harmonic series.
Step 1 · Radius of convergence
$$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac n{n+1} = 1 \implies R=1. $$
Step 2 · Endpoint $x=1$
$\sum_{n=1}^{\infty}(-1)^{n-1}/n=1-\frac12+\frac13-\cdots$ is the alternating harmonic series; by the Leibniz test it converges (to $\ln2$), and since $\sum1/n$ diverges, this convergence is conditional.
Step 3 · Endpoint $x=-1$
$$ \sum_{n=1}^{\infty}(-1)^{n-1}(-1)^n\frac1n = \sum_{n=1}^{\infty}(-1)^{2n-1}\frac1n = -\sum_{n=1}^{\infty}\frac1n, $$ the negative of the harmonic series, which diverges.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1$ |
| At $x=1$ | Converges conditionally, sum $=\ln2$ |
| At $x=-1$ | Diverges |
| Interval of convergence | $(-1,1]$ |
Additional Notes
- This series is precisely $\ln(1+x)$, the mirror image of Problem 225's $-\ln(1-x)=\sum x^n/n$: replacing $x\to-x$ swaps the roles of the two endpoints, exactly as seen by comparing this problem's interval $(-1,1]$ against Problem 225's $[-1,1)$.
- The closed form confirms the answer directly: $\ln(1+1)=\ln2$ is finite (matching convergence at $x=1$), while $\ln(1+(-1))=\ln0=-\infty$ is undefined (matching divergence at $x=-1$).
- This is the source identity behind the closed forms quoted in Problems 212 and 225 — the constant $\ln2=\sum(-1)^{n-1}/n$ recurs as one of the handful of "signature constants" running through this entire problem collection.