To solve this problem, we need to determine the number of terms $n$ required to ensure the error of the series:

$\sum_{n=1}^\infty \frac{(-1)^n}{n^{1.1}}$

is less than 0.001. This is an alternating series, and the error for such a series can be estimated by the magnitude of the first omitted term:

$| \text{error} | < |a_{n+1}| = \frac{1}{(n+1)^{1.1}}.$

Steps to Solve:

1. Set up the inequality for the error: $\frac{1}{(n+1)^{1.1}} < 0.001$

2. Solve for $n+1$: Take the reciprocal: $(n+1)^{1.1} > 1000$

3. Solve for $n+1$ using logarithms or exponentiation: $n+1 > 1000^{\frac{1}{1.1}}$

   Calculate $1000^{\frac{1}{1.1}}$: $1000^{\frac{1}{1.1}} \approx 373.69$

   So, $n+1 > 373.69 \quad \implies \quad n > 372.69.$

4. Conclusion: Since $n$ must be an integer, we round up: $n = 373.$

Thus, 373 terms are required to ensure the error is less than 0.001.

Would you like further details or a step-by-step explanation of any specific part of this calculation?

Related Questions:

1. How does the alternating series error estimate work?
2. What are the implications of the $1.1$ exponent in the series convergence rate?
3. How does rounding affect the accuracy of $n$?
4. Can a numerical approach verify the analytical result?
5. How would this change if the error bound were $0.0001$ instead?

Tip:

For alternating series, the error is always less than the magnitude of the first omitted term, making it easy to estimate accuracy.