Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{(n+2)(n+3)}$.
Explanation of Method
The ratio test gives radius $1$. Since the coefficient $b_n=\dfrac n{(n+2)(n+3)}\sim\dfrac1n$, direct limit comparison with the divergent harmonic series settles $x=1$ cleanly, while the same asymptotic combined with monotonicity settles $x=-1$ via the Leibniz test.
- Apply the ratio test to find $R$.
- Test $x=1$ via limit comparison with $\sum1/n$.
- Test $x=-1$ via the Leibniz test, and check absolute convergence.
Step 1 · Radius of convergence
$$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)(n+2)}{n(n+4)} = 1 \implies R=1. $$
Step 2 · Endpoint $x=1$: limit comparison
Let $b_n=\dfrac n{(n+2)(n+3)}$. Then $$ \lim_{n\to\infty}\frac{b_n}{1/n} = \lim_{n\to\infty}\frac{n^2}{(n+2)(n+3)} = 1, $$ a finite positive limit. Since $\sum1/n$ diverges, by the Limit Comparison Test, $\sum b_n=\sum_{n=1}^{\infty}\dfrac n{(n+2)(n+3)}$ also diverges.
Step 3 · Endpoint $x=-1$: Leibniz test
$b_n>0$; a short computation shows $b_{n+1}<b_n$ for $n\ge3$ (eventually decreasing, which suffices), and $b_n\sim1/n\to0$. By the Leibniz test, $\sum(-1)^nb_n$ converges at $x=-1$. Since $\sum|(-1)^nb_n|=\sum b_n$ diverges (Step 2), the convergence at $x=-1$ is conditional.
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1$ |
| At $x=1$ | Diverges (limit comparison with $\sum1/n$) |
| At $x=-1$ | Converges conditionally (Leibniz test) |
| Interval of convergence | $[-1,1)$ |
Additional Notes
- Limit comparison with the harmonic series is the cleanest way to settle boundary divergence whenever a coefficient is only asymptotically $\sim1/n$ (rather than exactly), avoiding any need to manipulate divergent series directly — the same technique underlies the Integral Test argument in Problem 232.
- This series shares its qualitative endpoint behavior with Problem 225's $\sum x^n/n$ (diverges at $+1$, converges conditionally at $-1$), since both coefficient sequences are asymptotically comparable to $1/n$.
- Compare with Problem 226, where an extra factor of $n$ in the denominator (giving coefficients $\sim1/n^2$) upgrades convergence at both endpoints to absolute — the boundary $1/n$ decay rate here is exactly the threshold separating these two outcomes.