Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}n$.
Explanation of Method
The ratio test gives radius $1$. At $x=1$ the series becomes the divergent harmonic series; at $x=-1$ it becomes the convergent (conditionally) alternating harmonic series — the classic asymmetric endpoint example.
- Apply the ratio test to find $R$.
- Test $x=1$: the harmonic series.
- Test $x=-1$: the alternating harmonic series, via the Leibniz test.
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 $x=1$
$\sum_{n=1}^{\infty}1/n$ is the harmonic series, which diverges.
Step 3 · Endpoint $x=-1$
$\sum_{n=1}^{\infty}(-1)^n/n$ is alternating with $1/n\downarrow0$, so by the Leibniz test it converges. Since $\sum|(-1)^n/n|=\sum1/n$ diverges, the convergence is conditional. Using the closed form $\sum x^n/n=-\ln(1-x)$: at $x=-1$, the sum equals $-\ln2$.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1$ |
| At $x=1$ | Diverges (harmonic series) |
| At $x=-1$ | Converges conditionally, sum $=-\ln2$ |
| Interval of convergence | $[-1,1)$ |
Additional Notes
- This is the archetypal "one endpoint in, one endpoint out" power series, and its closed form $-\ln(1-x)$ is the same function (up to sign) driving Problem 230's scaled version $\sum x^n/(na^n)$.
- Compare with Problem 228's series $\sum(-1)^{n-1}x^n/n=\ln(1+x)$: the two are mirror images of each other under $x\to-x$, so their surviving endpoints are swapped ($x=-1$ here vs. $x=1$ there).
- The harmonic-series divergence at $x=1$ recurs throughout this collection as the go-to counterexample for boundary divergence (see also Problems 226's telescoping contrast and 232's $\sum\ln n/n$).