Using Leibniz's test, prove the convergence of: sum
Using Leibniz's test, prove the convergence of: $$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}. $$
Explanation of Method
The sequence $b_n=\frac1{n(n+1)}$ is positive, decreasing, and tends to $0$, so Leibniz's test applies directly. Moreover, since $\sum\frac1{n(n+1)}$ itself telescopes to a finite value ($=1$, via partial fractions), the series actually converges absolutely, and the same telescoping trick evaluates the alternating sum exactly.
- Verify the three Leibniz conditions for $b_n=\frac1{n(n+1)}$.
- Show absolute convergence via the telescoping identity $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$.
- Evaluate the exact sum using the alternating telescoping partial sum.
Step 1 · Verify Leibniz's conditions
$b_n=\frac1{n(n+1)}>0$ for all $n\ge1$. Since $f(x)=\frac1{x(x+1)}$ has $f'(x)=-\frac{2x+1}{[x(x+1)]^2}<0$, $b_n$ is strictly decreasing. And $b_n=\frac1{n^2+n}\to0$. All conditions hold, so the series converges.
Step 2 · Absolute convergence via telescoping
Using $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$:
$$ \sum_{n=1}^{N}\frac1{n(n+1)} = \sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right) = 1-\frac1{N+1} \longrightarrow 1. $$
Since $\sum|a_n|=\sum\frac1{n(n+1)}=1<\infty$, the series converges absolutely.
Step 3 · Evaluate the exact sum
Using the same telescoping applied with the alternating sign:
$$ \sum_{n=1}^{N} \frac{(-1)^n}{n(n+1)} = \sum_{n=1}^{N} (-1)^n\left(\frac1n-\frac1{n+1}\right) = -1+\frac{(-1)^N}{N+1} \longrightarrow -1. $$
Summary Table
| Quantity | Value |
|---|---|
| $\sum\frac1{n(n+1)}$ (absolute series) | $1$ |
| $\sum\frac{(-1)^n}{n(n+1)}$ (this series) | $-1$ |
| Conclusion | Converges absolutely |
Additional Notes
- Unlike Problem 103's family $\sum(-1)^{n-1}/n^\alpha$, where only $\alpha>1$ gives absolute convergence, here the decay rate $\frac1{n(n+1)}\sim\frac1{n^2}$ is strong enough that Leibniz's test is actually unnecessary for establishing convergence — direct comparison with the convergent $p$-series ($\alpha=2$) would suffice.
- The telescoping structure makes this one of the rare alternating series problems where an exact closed-form sum ($-1$) is obtainable directly, rather than only a convergence classification.
- This telescoping technique — writing $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$ — is the same partial-fraction decomposition used throughout the early telescoping-series problems in this collection (Problems 2–5), here applied to an alternating rather than a plain series.
Result
The series converges absolutely, with sum $-1$: