Determine the radius of convergence and test convergence
Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}$.
Explanation of Method
The ratio test gives radius $1$. Since $1/(n(n+1))\sim1/n^2$, both endpoints converge absolutely by direct comparison — no need for the Leibniz test at $x=-1$, and the value at $x=1$ can be found exactly via the telescoping partial-fraction decomposition.
- Apply the ratio test to find $R$.
- Test $x=1$ via a telescoping sum, and $x=-1$ via absolute comparison to $\sum1/n^2$.
- Evaluate the sum at $x=-1$ using partial fractions.
Step 1 · Radius of convergence
$$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{n(n+1)}{(n+1)(n+2)} = \lim_{n\to\infty}\frac n{n+2} = 1 \implies R=1. $$
Step 2 · Endpoint $x=1$: telescoping
Using $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$: $$ \sum_{n=1}^{N}\left(\frac1n-\frac1{n+1}\right) = 1-\frac1{N+1} \longrightarrow 1. $$ So $\sum_{n=1}^{\infty}1/(n(n+1))=1$, converges (and absolutely, since all terms are positive).
Step 3 · Endpoint $x=-1$: absolute convergence and value
Since $1/(n(n+1))<1/n^2$ for all $n\ge1$ and $\sum1/n^2$ converges, both endpoints converge absolutely by direct comparison — no delicate alternating-series argument is needed. To find the value at $x=-1$: using $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$, $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)} = \sum_{n=1}^{\infty}\frac{(-1)^n}n - \sum_{n=1}^{\infty}\frac{(-1)^n}{n+1}. $$ The first sum is $-\ln2$ (Problem 225). Reindexing the second with $m=n+1$: $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n+1} = \sum_{m=2}^{\infty}\frac{(-1)^{m-1}}m = \left(\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}m\right) - 1 = \ln2-1. $$ So $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)} = -\ln2-(\ln2-1) = 1-2\ln2. $$
Summary Table
| Property | Value |
|---|---|
| Radius of convergence | $R=1$ |
| At $x=1$ | Converges absolutely, sum $=1$ |
| At $x=-1$ | Converges absolutely, sum $=1-2\ln2$ |
| Interval of convergence | $[-1,1]$ |
Additional Notes
- This is one of the few series in this batch whose interval of convergence is the full closed $[-1,1]$: the extra factor of $n$ in the denominator (compared to Problem 225's $\sum x^n/n$) is exactly enough to upgrade convergence at both endpoints from conditional/absent to fully absolute.
- The value $1$ at $x=1$ is exactly Problem 213's telescoping sum evaluated at $x=0$ shifted by an index — both problems exploit the identical partial-fraction identity $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$.
- The clean value $1$ at $x=1$ arises because the telescoping structure collapses to a finite constant, while the value $1-2\ln2\approx-0.386$ at $x=-1$ is a direct combination of two shifted copies of Problem 225's $\ln2$.