Show that the following series converges and calculate
Show that the following series converges and calculate its sum with accuracy $0.01$: $$ \sum_{n=2}^{\infty} \frac{(-1)^n}{n(n+1)(n+2)}. $$
Explanation of Method
With $a_n=\frac1{n(n+1)(n+2)}$ positive, decreasing, and $\to0$, Leibniz's test applies. Since the cubic-growth denominator decays extremely fast, the required accuracy is met after just two or three terms — far fewer than the pure $p$-series case of Problem 121.
- Verify convergence via Leibniz's test.
- Solve for the smallest $N$ with $(N+1)(N+2)(N+3)>100$.
- Compute the partial sum to the required accuracy.
Step 1 · Verify convergence
$a_n=\frac1{n(n+1)(n+2)}>0$, and since $n(n+1)(n+2)$ is strictly increasing, $a_n$ is decreasing; also $a_n\to0$. By Leibniz's test the series converges.
Step 2 · Determine the number of terms
Requiring $a_{N+1}=\frac1{(N+1)(N+2)(N+3)}<0.01$, i.e. $(N+1)(N+2)(N+3)>100$: at $N=2$, $(3)(4)(5)=60<100$ (insufficient); at $N=3$, $(4)(5)(6)=120>100$ (sufficient). So $N=3$ (two terms of the series starting from $n=2$) suffices.
Step 3 · Compute the partial sum
$$ S = \frac{(-1)^2}{2\cdot3\cdot4} + \frac{(-1)^3}{3\cdot4\cdot5} = \frac1{24}-\frac1{60} = \frac{5-2}{120} = \frac1{40} = 0.025. $$
The error is bounded by $a_4=\frac1{4\cdot5\cdot6}=\frac1{120}\approx0.0083<0.01$, confirming the accuracy.
Summary Table
| Quantity | Value |
|---|---|
| Terms needed | 2 (through $n=3$) |
| Partial sum | $\dfrac1{40}=0.025$ |
| Error bound | $\dfrac1{120}\approx0.0083<0.01$ |
Additional Notes
- Since $a_n\sim\frac1{n^3}$ for large $n$, this series' terms decay even faster than Problem 121's $1/n^4$-driven series relative to the number of terms needed — here just two terms suffice, versus three for Problem 121, illustrating how the specific polynomial growth rate of the denominator directly controls the practical cost of achieving a target accuracy.
- The triple-product denominator $n(n+1)(n+2)$ is also amenable to a partial-fraction telescoping decomposition (as in Problem 104's simpler $n(n+1)$ case), which could in principle yield a closed-form sum, though the direct numerical approach used here is simpler for the accuracy requirement at hand.
- Adding one more term ($n=4$) gives $S_4=\frac1{40}+\frac1{120}=\frac1{30}\approx0.0333$, illustrating how the alternating partial sums bracket the true value from both sides as more terms are included.
Result
The series converges, with sum accurate to $0.01$: