Let's analyze the convergence of each series in detail.

Problem Breakdown:

We are tasked to determine whether the given infinite series:

1. Converges absolutely,
2. Converges conditionally, or
3. Diverges.

Series:

1. $\sum_{n=1}^\infty \frac{3 + \sin(n)}{n(1 + e^n)}$,
2. $\sum_{n=1}^\infty \frac{(-1)^n 20}{e^n - e^{-n}}$,
3. $\sum_{n=1}^\infty \frac{n!}{n^n}$.

(i) $\sum_{n=1}^\infty \frac{3 + \sin(n)}{n(1 + e^n)}$:

Step 1: Examine the general term

The term is: $a_n = \frac{3 + \sin(n)}{n(1 + e^n)}.$

• $1 + e^n > e^n$, so: $a_n < \frac{3 + \sin(n)}{n e^n}.$

Step 2: Dominant behavior for large $n$

• For large $n$, $e^n$ grows exponentially, making the denominator grow much faster than the numerator. Thus: $a_n \sim \frac{1}{e^n}.$

Step 3: Convergence

The series $\sum \frac{1}{e^n}$ is a geometric series with ratio $r = \frac{1}{e} < 1$, which converges. Since $a_n$ is bounded above by a term of a convergent series, by the comparison test: $\sum_{n=1}^\infty \frac{3 + \sin(n)}{n(1 + e^n)} \text{ converges absolutely.}$

(ii) $\sum_{n=1}^\infty \frac{(-1)^n 20}{e^n - e^{-n}}$:

Step 1: Simplify the general term

The denominator is: $e^n - e^{-n} = e^n(1 - e^{-2n}).$ Thus, the term becomes: $a_n = \frac{(-1)^n 20}{e^n(1 - e^{-2n})}.$

Step 2: Examine the dominant term

For large $n$, $e^{-2n} \to 0$, so: $a_n \sim \frac{(-1)^n 20}{e^n}.$

Step 3: Alternating series test

• $b_n = \frac{20}{e^n}$ is positive, decreasing, and $\lim_{n \to \infty} b_n = 0$.
• By the Alternating Series Test, the series converges.

Step 4: Absolute convergence

For absolute convergence, consider $|a_n| = \frac{20}{e^n}$, which is a geometric series with $r = \frac{1}{e} < 1$. Hence, the series converges absolutely.

Conclusion: The series converges absolutely.

(iii) $\sum_{n=1}^\infty \frac{n!}{n^n}$:

Step 1: Examine the general term

The term is: $a_n = \frac{n!}{n^n}.$

Step 2: Behavior of $n!$

Using Stirling's approximation for $n!$, we have: $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$ Thus: $a_n \sim \frac{\sqrt{2\π n} \left(\frac{n}{e}\right)^n}{n^n} = \frac{\sqrt{2\pi n}}{e^n}.$

Step 3: Dominant term for large $n$

As $n \to \infty$, $e^n$ grows exponentially, dominating the numerator $\sqrt{2\pi n}$. Thus: $a_n \to 0 \text{ exponentially fast}.$

Step 4: Convergence

The series $\sum_{n=1}^\infty a_n$ involves terms decreasing faster than a geometric series. By the comparison test with a convergent geometric series (e.g., $\sum \frac{1}{e^n}$), this series converges.

Conclusion: The series converges absolutely.

Final Answers:

1. (i): Converges absolutely.
2. (ii): Converges absolutely.
3. (iii): Converges absolutely.

Would you like a more detailed explanation or further clarification?

Here are 5 related questions to expand your understanding:

1. What is the difference between absolute and conditional convergence?
2. How does the comparison test help analyze series convergence?
3. Can you apply the Ratio Test to any of these series?
4. How does the Alternating Series Test ensure conditional convergence?
5. What role does Stirling's approximation play in understanding factorial growth?

Tip: When analyzing a series, always consider the dominant term for large $n$ to predict the series' behavior effectively!