The given image contains the mathematical series:

$\sum_{n=0}^\infty \frac{(-1)^{n+6}n}{n^2 + 9}.$

Step 1: Analyze the Series

The general term of the series is: $a_n = \frac{(-1)^{n+6} n}{n^2 + 9}.$

The series alternates in sign due to the factor $(-1)^{n+6}$, where $n+6$ does not affect alternation (since it's equivalent to just $n$ in terms of the parity of the exponent).

Step 2: Convergence Check (Alternating Series Test)

To determine convergence, we need to check the following:

1. Monotonicity: The magnitude of the terms $\left|\frac{n}{n^2 + 9}\right|$ must decrease as $n$ increases.
2. Limit: The term $\frac{n}{n^2 + 9}$ must approach 0 as $n \to \infty$.

Monotonicity

The function $\frac{n}{n^2 + 9}$ is positive for $n \geq 0$. To test whether it decreases, compute the derivative of $f(n) = \frac{n}{n^2 + 9}$: $f'(n) = \frac{(n^2 + 9) \cdot 1 - n \cdot 2n}{(n^2 + 9)^2} = \frac{9 - n^2}{(n^2 + 9)^2}.$ For $n > 3$, $f'(n) < 0$, indicating the terms decrease for large $n$.

Limit of Terms

Compute the limit as $n \to \infty$: $\lim_{n \to \infty} \frac{n}{n^2 + 9} = \lim_{n \to \infty} \frac{1}{n + \frac{9}{n}} = 0.$

Since the terms decrease and approach 0, the alternating series converges.

Step 3: Further Simplification

The series does not telescope, and there is no closed-form simplification for the sum. However, it converges due to the alternating series test.

Let me know if you'd like more details or a numerical approximation of the sum!

Related Questions

1. What are the conditions for the convergence of an alternating series?
2. How is the alternating series remainder used to approximate sums?
3. Can the convergence rate of this series be determined?
4. How would the series change if the numerator was $(-1)^n n^2$?
5. What techniques exist for estimating sums of convergent series numerically?

Tip

When analyzing alternating series, always verify both monotonicity and the limit of terms to ensure convergence.