Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{(\ln x)^n}n. $$
Explanation of Method
With $y=\ln x$, this is again the classical logarithmic series $\sum y^n/n$, converging for $|y|<1$. Translating $|\ln x|<1$ back to $x$ (recalling $x>0$ is required for $\ln x$ to be defined) gives the interval $(1/e,e)$, with the same single conditionally-convergent endpoint mechanism as Problems 163 and 166.
- Substitute $y=\ln x$ and find the radius of convergence in $y$.
- Translate to an interval in $x$, respecting the domain $x>0$.
- Test both endpoints.
Step 1 · Radius of convergence in $y$
With $y=\ln x$: $\left|\dfrac{a_{n+1}}{a_n}\right|=|y|\dfrac n{n+1}\to|y|$, so $\sum y^n/n$ converges for $|y|<1$.
Step 2 · Translate to $x$
$|\ln x|<1\iff-1<\ln x<1\iff\dfrac1e<x<e$.
Step 3 · Test endpoints
At $x=e$ ($\ln x=1$): $\sum1/n$ diverges. At $x=1/e$ ($\ln x=-1$): $\sum(-1)^n/n$ converges (Leibniz) but not absolutely — conditionally convergent. The series equals $-\ln(1-\ln x)$ throughout its region of convergence.
Summary Table
| Region | Behavior |
|---|---|
| $x\in(1/e,e)$ | Absolutely convergent |
| $x=1/e$ | Conditionally convergent |
| $x\in(0,1/e)\cup[e,\infty)$ | Diverges |
Additional Notes
- This is the exponential-substitution twin of Problem 163: setting $y=\ln x$ instead of $y=2x$ transforms the interval of convergence from a symmetric interval around $0$ into a multiplicative interval $(1/e,e)$ around $x=1$.
- The domain restriction $x>0$ (needed for $\ln x$ to be defined) plays no additional role here beyond what's already implied by the convergence interval, since $(1/e,e)\subset(0,\infty)$ automatically.
- Just as in Problems 163 and 166, the conditionally convergent endpoint is exactly the one where the substituted variable equals $-1$, reflecting the alternating harmonic series' role as the universal boundary case for this family of logarithmic-type series.
Result
The convergence regions are as summarized above: