Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}n x^n$.
Explanation of Method
The ratio test gives radius $1$, since $\ln(n+1)/\ln n\to1$. At $x=1$ the Integral Test shows divergence, since $\int\ln x/x\,dx=(\ln x)^2/2\to\infty$. At $x=-1$, since $\ln n/n$ is eventually decreasing to $0$, the Leibniz test gives conditional convergence.
- Apply the ratio test to find $R=1$.
- Test $x=1$ via the Integral Test.
- Test $x=-1$ via the Leibniz test.
Step 1 · Radius of convergence
With $a_n=(\ln n)/n$: since $\ln(n+1)/\ln n\to1$ and $n/(n+1)\to1$: $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = 1 \implies R=1. $$
Step 2 · Endpoint $x=1$: Integral Test
$f(x)=\ln x/x$ is eventually positive, continuous, decreasing (for $x>e$). $$ \int_2^{\infty}\frac{\ln x}x\,dx = \left[\frac{(\ln x)^2}2\right]_2^{\infty} = \infty. $$ The integral diverges, so by the Integral Test, $\sum(\ln n)/n$ diverges.
Step 3 · Endpoint $x=-1$: Leibniz test
Let $b_n=(\ln n)/n$. For $f(t)=(\ln t)/t$: $f'(t)=(1-\ln t)/t^2<0$ for $t>e$, so $b_n$ is decreasing for $n\ge3$. Also $\lim b_n=0$ (by L'Hôpital, $\ln n/n\to0$). By the Leibniz test, $\sum(-1)^nb_n$ converges at $x=-1$. Since $\sum|b_n|=\sum(\ln n)/n$ diverges (Step 2), this convergence is conditional.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1$ |
| At $x=1$ | Diverges (Integral Test) |
| At $x=-1$ | Converges conditionally (Leibniz test) |
| Interval of convergence | $[-1,1)$ |
Additional Notes
- This series behaves qualitatively identically to Problem 225's $\sum x^n/n$ at both endpoints, even though $\ln n/n$ decays more slowly than $1/n$ — the logarithmic growth is too weak to change the threshold behavior, only the specific values of the resulting sums.
- The Integral Test computation $\int\ln x/x\,dx=(\ln x)^2/2$ is the same substitution ($u=\ln x$) used in Problem 200's convergence proof for $\sum1/(n\ln^2n)$; here the exponent on $\ln x$ in the denominator is one less, which is exactly the difference between convergence and divergence of the corresponding integral.
- Unlike Problem 227, where a limit comparison test settled the boundary directly, here the coefficient $\ln n/n$ has no simpler comparable series with a known closed-form ratio, so the Integral Test is genuinely the natural tool.