Test the conditional, absolute, and uniform convergence
Test the conditional, absolute, and uniform convergence of $\displaystyle\sum_{n=1}^{\infty}\left(\frac{x}{x+1}\right)^{\alpha n}$, $\alpha\in\mathbb R$.
Explanation of Method
This is a geometric series with ratio $r(x)=\left(\dfrac{x}{x+1}\right)^{\alpha}$, converging exactly when $|r(x)|<1$. Since $\left|\dfrac x{x+1}\right|<1\iff|x|<|x+1|\iff x>-\tfrac12$ (by squaring both sides), the sign of $\alpha$ simply decides whether the convergence region is $x>-\tfrac12$ or $x<-\tfrac12$. Because a geometric series never converges conditionally, absolute and ordinary convergence regions always coincide.
- Solve $|x|<|x+1|\iff x>-\tfrac12$ by squaring.
- Case on the sign of $\alpha$ to find the convergence region.
- Note absolute/ordinary convergence coincide, and locate uniform convergence on compact subsets.
Step 1 · The key inequality
For $x\ne-1$: $\left|\dfrac x{x+1}\right|<1 \iff x^2<(x+1)^2=x^2+2x+1 \iff x>-\frac12$ (squaring is valid since both sides are non-negative). Likewise $\left|\dfrac x{x+1}\right|>1\iff x<-\frac12$ (and $x\ne-1$).
Step 2 · Case on $\alpha$
$\alpha>0$: the series converges iff $\left|\dfrac x{x+1}\right|^\alpha<1\iff\left|\dfrac x{x+1}\right|<1\iff x>-\dfrac12$. So the region is $\left(-\dfrac12,\infty\right)$ (note $x=-1$ is already excluded, since $-1<-\tfrac12$).
$\alpha<0$: writing $\beta=-\alpha>0$, convergence needs $\left|\dfrac x{x+1}\right|^{-\beta}<1\iff\left|\dfrac x{x+1}\right|>1\iff x<-\dfrac12$. Excluding the undefined point $x=-1$, the region is $(-\infty,-1)\cup(-1,-\tfrac12)$.
$\alpha=0$: every term equals $1$, so the series diverges for all $x$.
Step 3 · Absolute and uniform convergence
A geometric series either converges absolutely or diverges — it is never conditionally convergent, since $|r^n|=|r|^n$ shares the exact convergence threshold with $r^n$. So the regions of absolute and ordinary convergence are identical in every case above. For uniform convergence: on any compact $[a,b]$ strictly inside the convergence region (so that $\sup_{[a,b]}|r(x)|=\rho<1$), the Weierstrass M-test with $M_n=\rho^{\alpha n}$ (or $\rho^{|\alpha| n}$ as appropriate) gives uniform convergence; near the boundary point $x=-\tfrac12$ or as $|x|\to\infty$, convergence is not uniform, since $r(x)\to1^-$ there.
Summary Table
| Case | Convergence region |
|---|---|
| $\alpha>0$ | $\left(-\dfrac12,\infty\right)$ |
| $\alpha<0$ | $(-\infty,-1)\cup\left(-1,-\dfrac12\right)$ |
| $\alpha=0$ | Diverges everywhere |
Additional Notes
- The threshold $x=-\tfrac12$ is exactly the fixed point where $\left|\dfrac x{x+1}\right|=1$, i.e. where $x$ and $x+1$ have equal magnitude — a clean algebraic condition that replaces messy case-by-case sign analysis with a single squaring step.
- Since $\alpha>0$ and $\alpha<0$ give regions on opposite sides of $x=-\tfrac12$, this problem is a nice illustration of how inverting the ratio ($r\to1/r$) via a sign flip in the exponent exactly reflects the convergence region across the critical point.
- As with every geometric series in this collection (compare Problems 184 and 185's Cauchy-criterion tests for $\sum x^n$), the absence of conditional convergence is a structural feature: it is only alternating or otherwise sign-varying series (as in Problem 201) that can converge without converging absolutely.
Result
Convergence regions determined by the sign of $\alpha$; absolute and ordinary convergence always coincide: