We are tasked with working on an alternating series given by the expression:

$\sum_{n=1}^{\infty} \frac{(-1)^n \cdot 400}{n^{0.9}}$

Part [A]: Calculate $S_3$

We need to calculate the sum of the first 3 terms of the series, $S_3$. The general term of the series is:

$a_n = \frac{(-1)^n \cdot 400}{n^{0.9}}$

The first three terms are given by:

1. $a_1 = \frac{(-1)^1 \cdot 400}{1^{0.9}} = \frac{-400}{1} = -400$
2. $a_2 = \frac{(-1)^2 \cdot 400}{2^{0.9}} = \frac{400}{2^{0.9}}$
3. $a_3 = \frac{(-1)^3 \cdot 400}{3^{0.9}} = \frac{-400}{3^{0.9}}$

Now, let's calculate the actual values for $a_2$ and $a_3$ and then sum the three terms.

$a_2 = \frac{400}{2^{0.9}} = \frac{400}{1.86607} \approx 214.366$ $a_3 = \frac{-400}{3^{0.9}} = \frac{-400}{2.74596} \approx -145.755$

Now, summing these up:

$S_3 = a_1 + a_2 + a_3 = -400 + 214.366 - 145.755 \approx -331.389$

So, the sum of the first three terms is:

$S_3 \approx -331.389$

Part [B]: Error Bound

For an alternating series, the error bound when approximating the sum by the first $n$ terms is given by the absolute value of the next term $|a_{n+1}|$. So, the error bound after three terms is $|a_4|$.

Let's calculate $a_4$:

$a_4 = \frac{400}{4^{0.9}} = \frac{400}{3.73719} \approx 107.093$

Therefore, the error bound is:

$\text{Error} \leq |a_4| \approx 107.093$

Final Answers:

• $S_3 \approx -331.389$
• $\text{Error} \leq 107.093$

Do you want more details on any of the steps?

Here are 5 related questions to deepen understanding:

1. How do you determine whether an alternating series converges?
2. What is the Alternating Series Test and how does it apply here?
3. Can you calculate $S_4$ and compare it to $S_3$ to see how the sum is changing?
4. How would you calculate the total sum of the series as $n \to \infty$?
5. Why is the error bound based on the next term in alternating series?

Tip: For alternating series, the error bound after $n$ terms is always less than or equal to the magnitude of the next term, which is a great way to estimate the accuracy of partial sums.