To solve the problem, we must determine how many terms are required in the given series:

$\sum_{n=1}^\infty \frac{3(-1)^n}{n^{1.2}}$

such that the error (remainder) is less than $0.001$.

Key Concepts:

1. Alternating Series Approximation:

   • For alternating series, the error of the partial sum $S_N$ (sum up to $N$-th term) is less than or equal to the absolute value of the first omitted term: $|R_N| \leq \left| a_{N+1} \right|$
   • Here, $a_n = \frac{3}{n^{1.2}}$, so $a_{N+1} = \frac{3}{(N+1)^{1.2}}$.

2. Condition:

   • Find $N$ such that: $\frac{3}{(N+1)^{1.2}} < 0.001$

Solve for $N$:

Rearrange the inequality to find $N+1$: $(N+1)^{1.2} > \frac{3}{0.001} = 3000$ $N+1 > 3000^{\frac{1}{1.2}}$ $N+1 > 3000^{0.8333}$

Now compute $3000^{0.8333}$: $3000^{0.8333} \approx 947.43$ Thus: $N+1 > 947.43 \quad \Rightarrow \quad N > 946.43$

Since $N$ must be an integer, we round up: $N = 947$

Final Answer:

To ensure the error is less than $0.001$, at least 947 terms should be used.

Let me know if you would like me to explain or verify any part of this further!

Related Questions:

1. What is the formula for the remainder of a general alternating series?
2. How can we verify that the series converges?
3. What is the behavior of the series as $n \to \infty$?
4. What happens if the exponent $1.2$ changes to another value?
5. Can we estimate the error for a different tolerance level, say $0.01$?

Tip: Always round up when determining the number of terms in error-bound problems, as fractional terms don't count.