To solve this, the series provided is alternating, and the problem requires determining the number of terms $n$ needed to ensure that the error is less than $0.01$.

For an alternating series, the error can be approximated by the magnitude of the next term: $\text{Error} \approx \left| a_{n+1} \right| < 0.01$

The general term of the series is: $a_n = \frac{(-1)^n (n^5)}{n^{0.6}}$

Simplified (ignoring the alternating factor): $a_n = \frac{n^5}{n^{0.6}} = n^{5 - 0.6} = n^{4.4}$

Steps:

1. Set $|a_{n+1}| < 0.01$: $\frac{(n+1)^5}{(n+1)^{0.6}} < 0.01 \implies (n+1)^{4.4} < 0.01$

2. Solve for $n+1$ by taking logarithms: $n+1 < \left(0.01\right)^{\frac{1}{4.4}}$

3. Compute $n+1$ and round up to the nearest integer.

Let me calculate this for you.import math

Solve for n+1 using the formula: n+1 < (0.01)^(1/4.4)

error_threshold = 0.01 exponent = 4.4

n_plus_1 = (error_threshold) ** (1 / exponent)

Find n by subtracting 1 and taking the ceiling for accuracy

n = math.ceil(n_plus_1 - 1) n00successThe calculated $n$ is $0$, meaning the series converges quickly, and even the first term would provide sufficient accuracy to meet the error threshold of $0.01$.

Would you like to see the detailed steps or have further clarifications? Here are some related questions to deepen your understanding:

1. How does the alternating series error estimation method work?
2. What would happen if the error threshold were reduced further?
3. Can we generalize this approach to other series?
4. How does the behavior of $n^{4.4}$ affect the series convergence?
5. How does rounding up influence the practical result?

Tip: For alternating series, focusing on the magnitude of the next term is a quick way to estimate the error.