Let's check whether the given series:

$U_n = \frac{n \cdot 2^n}{(n+1) \cdot 3^n}$

is convergent using the D'Alembert Ratio Test. The ratio test states that for a series $\sum a_n$, if:

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

• If $L < 1$, the series converges.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

Step 1: Find the ratio $\frac{a_{n+1}}{a_n}$

For the given $U_n = \frac{n \cdot 2^n}{(n+1) \cdot 3^n}$, the next term is $U_{n+1}$:

$U_{n+1} = \frac{(n+1) \cdot 2^{n+1}}{(n+2) \cdot 3^{n+1}}$

Now, compute the ratio:

$\frac{U_{n+1}}{U_n} = \frac{\frac{(n+1) \cdot 2^{n+1}}{(n+2) \cdot 3^{n+1}}}{\frac{n \cdot 2^n}{(n+1) \cdot 3^n}}$

Simplifying this expression:

$\frac{U_{n+1}}{U_n} = \frac{(n+1) \cdot 2^{n+1} \cdot (n+1) \cdot 3^n}{(n+2) \cdot 3^{n+1} \cdot n \cdot 2^n}$

$= \frac{(n+1)^2 \cdot 2 \cdot 3^n}{(n+2) \cdot 3^{n+1} \cdot n}$

$= \frac{(n+1)^2 \cdot 2}{(n+2) \cdot 3 \cdot n}$

Step 2: Take the limit as $n \to \infty$

Now, compute the limit of this expression as $n \to \infty$:

$L = \lim_{n \to \infty} \frac{(n+1)^2 \cdot 2}{(n+2) \cdot 3 \cdot n}$

As $n \to \infty$, $(n+1) \approx n$ and $(n+2) \approx n$, so the expression simplifies to:

$L = \lim_{n \to \infty} \frac{n^2 \cdot 2}{n \cdot 3 \cdot n}$

$L = \lim_{n \to \infty} \frac{2n^2}{3n^2} = \frac{2}{3}$

Step 3: Apply the Ratio Test

Since $L = \frac{2}{3}$, which is less than 1, the series converges by the D'Alembert Ratio Test.

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Do you have any questions about the solution? Would you like further details?

Here are some related questions to explore further:

1. How can the Ratio Test be used for series involving exponential functions?
2. What happens if $L = 1$ in the Ratio Test? How do you proceed then?
3. How does the Root Test compare with the Ratio Test for determining convergence?
4. Can we apply the Ratio Test to series with factorial terms?
5. What are some examples of series where the Ratio Test does not give a conclusive result?

Tip: When dealing with exponential functions in series, always try to isolate terms involving the base of the exponentials to simplify the limit calculation.