The problem asks to determine the number of terms $n$ that must be added in the series to achieve an error less than 5.

Series Expression:

$\sum_{n=1}^{\infty} \frac{(-1)^n 300}{n^{0.6}}$ where the error condition is $|\text{error}| < 5$.

Step-by-Step Approach:

1. Understanding the Series: This is an alternating series, and its general term is: $a_n = \frac{(-1)^n 300}{n^{0.6}}$ For alternating series, the error after adding $n$ terms is approximately the absolute value of the first omitted term.

2. Error Condition: The first omitted term (the next term after the $n$-th term) is: $a_{n+1} = \frac{300}{(n+1)^{0.6}}$ The error condition requires: $\left| \frac{300}{(n+1)^{0.6}} \right| < 5$

3. Solve for $n$: Solving the inequality for $n$, $\frac{300}{(n+1)^{0.6}} < 5$ Simplifying this, $(n+1)^{0.6} > \frac{300}{5} = 60$ Raising both sides to the power of $\frac{1}{0.6}$, $n+1 > 60^{\frac{1}{0.6}} = 60^{\frac{5}{3}} \approx 132.6$ Therefore, $n > 131.6$ So the smallest integer $n$ that satisfies this condition is $n = 132$.

Conclusion:

To ensure that the error is less than 5, you must sum at least 132 terms of the series. Therefore, $n = 35$ in the image is too small.

Would you like more details on any of the steps?

Here are 5 related questions you might find interesting:

1. How do you estimate the error in an alternating series?
2. How does the alternating series test guarantee convergence?
3. What would the error be if you summed only 35 terms?
4. How would this change if the exponent in the denominator was different?
5. What if the numerator's constant was different?

Tip: Always remember that for alternating series, the error is typically dominated by the first omitted term after partial sums.