To analyze the convergence of the sequence $a_n = \frac{n^n}{(n+1)^n}$ using the D'Alembert's Ratio Test, we need to compute the limit:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$

Step 1: Calculate $\frac{a_{n+1}}{a_n}$

$\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^{n+1}}{(n+2)^{n+1}}}{\frac{n^n}{(n+1)^n}} = \frac{(n+1)^{n+1}}{(n+2)^{n+1}} \cdot \frac{(n+1)^n}{n^n}$

Step 2: Simplify the expression

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1} \cdot (n+1)^n}{(n+2)^{n+1} \cdot n^n}$

Now simplify the powers of $(n+1)$:

$= \frac{(n+1)^{2n+1}}{(n+2)^{n+1} \cdot n^n}$

We can separate this into two parts:

$= \frac{(n+1)^{2n+1}}{(n+2)^{n+1}} \cdot \frac{1}{n^n}$

Step 3: Compute the limit as $n \to \infty$

At large $n$, we focus on the dominant behavior of each term. Let's break it down:

1. $\frac{(n+1)^{2n+1}}{(n+2)^{n+1}}$ behaves like $1$ because as $n \to \infty$, $n+1$ and $n+2$ are approximately equal.
2. The term $\frac{1}{n^n}$ approaches 0 as $n \to \infty$, since it grows very quickly.

Thus:

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0$

Step 4: Conclusion

Since the ratio test gives $L = 0$, which is less than 1, the series converges.

Would you like further details on any part of this process?

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Here are five related questions you might explore next:

1. How does the ratio test compare to other convergence tests?
2. What is the general form of D'Alembert's ratio test for infinite series?
3. Can we apply the root test to the same sequence? If so, what do we find?
4. What happens to the sequence $\frac{n^n}{(n+1)^n}$ for small $n$?
5. What other convergence tests are useful for factorial-like expressions?

Tip: When applying the ratio test, ensure the limit exists and simplifies correctly; this will give you a definitive conclusion on convergence.