Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac n{x^n}. $$
Explanation of Method
Writing $y=1/x$, this is the power series $\sum ny^n$, resolved directly by the ratio test. Since the series of absolute values shares the exact same convergence criterion as the original (no sign alternation occurs for $x>0$, and even for $x<0$ the terms $n/x^n$ don't tend to $0$ at the boundary), there is no region of conditional convergence at all.
- Rewrite as a power series in $y=1/x$ and apply the ratio test.
- Test both endpoints $x=\pm1$.
- Confirm there is no conditional-convergence region, since the absolute series shares the same threshold.
Step 1 · Ratio test
With $y=1/x$: $\left|\dfrac{(n+1)y^{n+1}}{ny^n}\right|\to|y|=1/|x|$, so the series converges for $1/|x|<1$, i.e. $|x|>1$, and diverges for $|x|<1$.
Step 2 · Test the endpoints
At $x=1$: $\sum n$ diverges (terms $\to\infty$). At $x=-1$: $\sum n(-1)^n$ has terms that oscillate in magnitude toward $\infty$, so $n(-1)^n\not\to0$; the series diverges by the $n$-th term test.
Step 3 · No conditional convergence
The series of absolute values $\sum n/|x|^n$ obeys the identical ratio-test threshold $|x|>1$, so wherever the original series converges, it converges absolutely — there is no region of purely conditional convergence. For $|x|>1$: $$ \sum_{n=1}^{\infty}\frac n{x^n} = \frac{1/x}{(1-1/x)^2} = \frac x{(x-1)^2}. $$
Summary Table
| Region | Behavior |
|---|---|
| $|x|>1$ | Absolutely convergent, sum $=x/(x-1)^2$ |
| $|x|\le1$ | Diverges |
Additional Notes
- This series is the direct evaluation-target family from Problem 143's differentiated-geometric-series technique, applied here symbolically at $y=1/x$ rather than at a specific numeric value like $\frac12$.
- The complete absence of conditional convergence is typical for power series with polynomially-growing coefficients (here, coefficient $n$): unlike Problem 163's $1/n$-weighted series (whose slow decay allows Leibniz's test to rescue one boundary point), the growing weight $n$ here ensures the terms fail to vanish at either endpoint.
- The closed form $x/(x-1)^2$ is the same rational function arising from differentiating the geometric series $\sum(1/x)^n=1/(x-1)$ (up to sign and shift), directly generalizing the numeric identity in Problem 143.
Result
Converges absolutely for $|x|>1$, with no conditional convergence: