Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{na^n}$, $a>0$.
Explanation of Method
This is Problem 225's series $\sum y^n/n$ with $y=x/a$ rescaled, so its behavior transfers directly: radius $R=a$, divergence at $x=a$ (harmonic series), conditional convergence at $x=-a$ (alternating harmonic series).
- Apply the ratio test to find $R=a$.
- Test $x=a$: reduces to the harmonic series.
- Test $x=-a$: reduces to the alternating harmonic series.
Step 1 · Radius of convergence
With $b_n=1/(na^n)$: $$ \lim_{n\to\infty}\left|\frac{b_{n+1}}{b_n}\right| = \lim_{n\to\infty}\frac{n}{(n+1)a} = \frac1a \implies R=a. $$
Step 2 · Endpoint $x=a$
$\sum_{n=1}^{\infty}\dfrac{a^n}{na^n}=\sum_{n=1}^{\infty}\dfrac1n$, the harmonic series, diverges.
Step 3 · Endpoint $x=-a$
$\sum_{n=1}^{\infty}\dfrac{(-a)^n}{na^n}=\sum_{n=1}^{\infty}\dfrac{(-1)^n}n$ converges conditionally by the Leibniz test (Problem 225), to the value $-\ln2$.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=a$ |
| At $x=a$ | Diverges (harmonic series) |
| At $x=-a$ | Converges conditionally, sum $=-\ln2$ |
| Interval of convergence | $[-a,a)$ |
Additional Notes
- This is Problem 225 in disguise: substituting $y=x/a$ turns this series into exactly $\sum y^n/n$, so every conclusion — radius, endpoint behavior, and the value $-\ln2$ — transfers immediately via the scaling $x=ay$.
- The closed form is $f(x)=-\ln(1-x/a)=\ln\!\left(\dfrac a{a-x}\right)$, which blows up logarithmically as $x\to a^-$, consistent with the harmonic-series divergence found at that endpoint.
- The parameter $a$ acts purely as a rescaling of the $x$-axis: it shifts the radius and both endpoints proportionally, but changes nothing about which endpoint converges — this is a common pattern whenever a family of series is built by substituting $x\to x/a$ into a fixed reference series.