The given series is:

$\left(\frac{1}{3}\right)^2 + \left(\frac{1 \cdot 2}{3 \cdot 5}\right)^2 + \left(\frac{1 \cdot 2 \cdot 3}{3 \cdot 5 \cdot 7}\right)^2 + \dots$

To analyze the convergence or divergence of this series, let's define the general term of the series, $a_n$:

$a_n = \left(\frac{1 \cdot 2 \cdot 3 \cdot \dots \cdot n}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)}\right)^2$

We are dealing with a series of the form:

$\sum_{n=1}^{\infty} a_n$

Ratio Test

One common method to test for convergence is the Ratio Test. We compute the limit:

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

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

Let's compute the ratio:

$\frac{a_{n+1}}{a_n} = \frac{\left(\frac{1 \cdot 2 \cdot 3 \cdot \dots \cdot (n+1)}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1) \cdot (2n+3)}\right)^2}{\left(\frac{1 \cdot 2 \cdot 3 \cdot \dots \cdot n}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)}\right)^2}$

This simplifies to:

$\frac{a_{n+1}}{a_n} = \left(\frac{n+1}{2n+3}\right)^2$

Now, let's find the limit as $n$ approaches infinity:

$L = \lim_{n \to \infty} \left(\frac{n+1}{2n+3}\right)^2 = \lim_{n \to \infty} \left(\frac{n(1 + \frac{1}{n})}{2n(1 + \frac{3}{2n})}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Since $L = \frac{1}{4} < 1$, the series converges by the Ratio Test.

Would you like more details or have any other questions?

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Here are 5 related questions:

1. What is the general form of the Ratio Test for series convergence?
2. How does the Root Test compare to the Ratio Test in terms of determining series convergence?
3. Can you find the sum of the convergent series?
4. What are some examples of divergent series and how do they differ from this one?
5. How would the convergence behavior change if the denominator had a different pattern?

Tip: The Ratio Test is particularly useful for series involving factorials or products, as it easily handles the growth rate of terms.