Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac1{n(x+2)^n}. $$
Explanation of Method
Substituting $y=1/(x+2)$ reduces this to the classical logarithmic series $\sum y^n/n$, whose radius of convergence is $1$ and whose single conditionally-convergent boundary point is $y=-1$. Translating back to $x$ gives an asymmetric two-piece convergence region with one genuine conditional-convergence point.
- Substitute $y=1/(x+2)$ and find the radius of convergence in $y$.
- Translate the condition $|y|<1$ back into a condition on $x$.
- Test both endpoints and identify the series as a logarithm.
Step 1 · Radius of convergence in $y$
With $y=1/(x+2)$: $\left|\dfrac{a_{n+1}}{a_n}\right|=|y|\dfrac n{n+1}\to|y|$, so $\sum y^n/n$ converges for $|y|<1$, diverges for $|y|>1$.
Step 2 · Translate to $x$
$|y|<1\iff|x+2|>1\iff x>-1$ or $x<-3$.
Step 3 · Test endpoints
At $x=-1$ ($y=1$): $\sum1/n$ diverges. At $x=-3$ ($y=-1$): $\sum(-1)^n/n$ converges (Leibniz) but not absolutely — conditionally convergent. The series equals $\ln\left(\dfrac{x+2}{x+1}\right)$ wherever it converges.
Summary Table
| Region | Behavior |
|---|---|
| $x\in(-\infty,-3)\cup(-1,\infty)$ | Absolutely convergent |
| $x=-3$ | Conditionally convergent |
| $x\in(-3,-1]$ | Diverges |
Additional Notes
- This is a horizontally shifted and reciprocated version of Problem 163's series: the same underlying logarithmic series $\sum y^n/n$ appears, but the substitution $y=1/(x+2)$ (rather than $y=2x$) turns the interior of convergence into the exterior of an interval in $x$.
- The single conditionally convergent point $x=-3$ traces directly back to the alternating harmonic series' convergence, the same mechanism underlying Problem 163's endpoint $x=\frac12$.
- The excluded gap $(-3,-1)$ corresponds to $|y|>1$, where the reciprocated substitution makes the "interior" of the $x+2$ disk the divergent region — a useful reminder that a reciprocal substitution can invert which side of a boundary is convergent.
Result
The convergence regions are as summarized above: