To determine if the series

$\sum_{n=1}^{\infty} (-1)^n \left( \frac{\sqrt{n}}{n+8} \right)$

is convergent or divergent using the Alternating Series Test, we need to check the following two conditions:

1. The terms decrease monotonically (in absolute value):
   We need to check if the sequence $a_n = \frac{\sqrt{n}}{n+8}$ is decreasing for all $n$ sufficiently large.

2. The limit of the terms as $n \to \infty$ is zero:
   We need to check if $\lim_{n \to \infty} a_n = 0$.

Step 1: Check if the terms go to 0 as $n \to \infty$

Consider the behavior of $a_n = \frac{\sqrt{n}}{n+8}$ as $n \to \infty$:

= \lim_{n \to \infty} \frac{\frac{\sqrt{n}}{n}}{1 + \frac{8}{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0.$$ Since the limit is 0, the first condition of the Alternating Series Test is satisfied. ### Step 2: Check if the sequence is monotonically decreasing We need to check if $$a_n = \frac{\sqrt{n}}{n+8}$$ is decreasing for large $$n$$. To verify this, we can compute the derivative of the function $$f(x) = \frac{\sqrt{x}}{x+8}$$ and check if it is negative for $$x \geq 1$$. First, calculate the derivative $$f'(x)$$ using the quotient rule: $$f'(x) = \frac{(x+8) \cdot \frac{1}{2\sqrt{x}} - \sqrt{x} \cdot 1}{(x+8)^2} = \frac{\frac{x+8}{2\sqrt{x}} - \sqrt{x}}{(x+8)^2}.$$ Simplify the numerator: $$\frac{x+8}{2\sqrt{x}} - \sqrt{x} = \frac{x+8 - 2x}{2\sqrt{x}} = \frac{8-x}{2\sqrt{x}}.$$ So, $$f'(x) = \frac{8-x}{2\sqrt{x}(x+8)^2}.$$ For $$x \geq 1$$, the numerator $$8 - x$$ is positive when $$x$$ is less than 8 and negative when $$x$$ is greater than 8. This implies that $$f'(x)$$ is positive when $$x < 8$$ and negative when $$x > 8$$. Therefore, the function $$f(x) = \frac{\sqrt{x}}{x+8}$$ is **decreasing** for $$x \geq 8$$, which is sufficient to conclude that $$a_n$$ decreases for large $$n$$. ### Conclusion Since both conditions of the **Alternating Series Test** are satisfied: 1. $$\lim_{n \to \infty} a_n = 0$$, 2. $$a_n$$ is eventually decreasing, the series $$\sum_{n=1}^{\infty} (-1)^n \left( \frac{\sqrt{n}}{n+8} \right)$$ **converges** by the Alternating Series Test. --- Would you like further details on this, or do you have another question? --- **Related questions**: 1. What is the Alternating Series Test, and when should it be applied? 2. How do you compute the derivative of a function using the quotient rule? 3. What is the behavior of the sequence $$\frac{\sqrt{n}}{n+8}$$ as $$n$$ approaches infinity? 4. How do you determine if a sequence is monotonic? 5. How can you check for convergence of a series using the Ratio Test or Root Test? **Tip**: Always ensure to simplify terms carefully when evaluating limits or checking if terms go to zero as $$n$$ approaches infinity.