Show that the following series converges and calculate
Show that the following series converges and calculate its sum with accuracy $0.01$: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4}. $$
Explanation of Method
Leibniz's test immediately gives convergence. The standard alternating-series error bound $|S-S_N|\le a_{N+1}$ then converts the accuracy requirement directly into a condition on $N$: solving $\frac1{(N+1)^4}<0.01$ tells us exactly how many terms are needed, after which the partial sum itself is the answer.
- Verify convergence via Leibniz's test.
- Solve $\frac1{(N+1)^4}<0.01$ for the required number of terms $N$.
- Compute the partial sum $S_N$ to the required accuracy.
Step 1 · Verify convergence
With $a_n=\frac1{n^4}$: positive, decreasing, and $\to0$. By Leibniz's test the series converges.
Step 2 · Determine the number of terms
The error satisfies $|S-S_N|\le a_{N+1}=\frac1{(N+1)^4}$. Requiring this $<0.01$: $(N+1)^4>100\Rightarrow N+1>\sqrt[4]{100}=\sqrt{10}\approx3.162$, so $N+1\ge4$, i.e. $N\ge3$. Checking: $a_4=\frac1{256}\approx0.0039<0.01$ (sufficient), while $a_3=\frac1{81}\approx0.0123>0.01$ (not sufficient) — confirming $N=3$ is the minimum.
Step 3 · Compute $S_3$
$$ S_3 = 1-\frac1{16}+\frac1{81} = \frac{1296-81+16}{1296} = \frac{1231}{1296} \approx 0.9498. $$
The error is bounded by $a_4=\frac1{256}\approx0.0039<0.01$, confirming the required accuracy.
Summary Table
| Quantity | Value |
|---|---|
| Terms needed | $N=3$ |
| $S_3$ | $\dfrac{1231}{1296}\approx0.9498$ |
| Error bound | $a_4\approx0.0039<0.01$ |
Additional Notes
- The alternating-series error bound $|S-S_N|\le a_{N+1}$ is a remarkably efficient tool: for a rapidly decaying sequence like $1/n^4$, only a handful of terms are needed for even fairly tight accuracy, in sharp contrast to the million-plus terms Problem 123 requires for a slowly decaying $1/\sqrt{n^2+1}$-type series.
- The exact value of this series is known in closed form: $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^4}=\frac{7\pi^4}{720}\approx0.9470$, and our three-term approximation $0.9498$ indeed lies within $0.01$ of this true value.
- This problem exemplifies the general strategy of Problems 121–124: convert a required accuracy into a bound on $N$ via the Leibniz error estimate, then simply compute that many terms — the harder problems in this group differ only in how slowly $a_n\to0$, which controls how large $N$ must be.
Result
Three terms suffice for accuracy $0.01$: