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The question asks to find the radius of convergence for the given series:

\[
\sum_{n=1}^\infty \frac{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)}{(2n)!} x^n
\]

### Step 1: Analyzing the series
The numerator \(2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)\) can be rewritten as \(2^n \cdot n!\) since it is the product of the first \(n\) even numbers. Substituting this into the series, the \(n\)-th term becomes:

\[
a_n = \frac{2^n \cdot n!}{(2n)!}.
\]

Thus, the series becomes:

\[
\sum_{n=1}^\infty \frac{2^n \cdot n!}{(2n)!} x^n.
\]

### Step 2: Using the Ratio Test to find the radius of convergence
The Ratio Test involves evaluating:

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.
\]

We calculate \( \frac{a_{n+1}}{a_n} \) as follows:
\[
a_{n+1} = \frac{2^{n+1} \cdot (n+1)!}{(2(n+1))!} = \frac{2^{n+1} \cdot (n+1)!}{(2n+2)!},
\]
\[
a_n = \frac{2^n \cdot n!}{(2n)!}.
\]

Dividing \(a_{n+1}\) by \(a_n\):

\[
\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1} \cdot (n+1)!}{(2n+2)!}}{\frac{2^n \cdot n!}{(2n)!}}
= \frac{2^{n+1} \cdot (n+1)! \cdot (2n)!}{2^n \cdot n! \cdot (2n+2)!}.
\]

Simplify the terms:
\[
\frac{a_{n+1}}{a_n} = 2 \cdot \frac{(n+1)}{(2n+2)(2n+1)}.
\]

As \(n \to \infty\), the dominant term in the denominator is \(4n^2\), so:
\[
\frac{a_{n+1}}{a_n} \approx \frac{2(n+1)}{4n^2} = \frac{2}{4n} = \frac{1}{2n}.
\]

Thus, the limit becomes:
\[
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0.
\]

### Step 3: Radius of Convergence
Since the ratio test shows the series converges for all \(x\), the radius of convergence is:

\[
R = \infty.
\]

Would you like further clarification on this process? Here are 5 related questions:

1. How do you determine the factorial of even numbers like \( (2n)! \)?
2. What is the purpose of the Ratio Test for finding convergence?
3. Can this series converge for all real numbers \(x\)?
4. How do we simplify products like \(2^n \cdot n!\)?
5. Why does the limit approaching 0 imply an infinite radius of convergence?

**Tip**: When dealing with factorial expressions, always try simplifying the largest term first to reduce complexity.

Find the radius of convergence for the series ∑ (2 ⋅ 4 ⋅ 6 ⋅ … ⋅ (2n)) / (2n)! ⋅ x^n.

Step-by-step solution for finding the radius of convergence of the series ∑ (2 ⋅ 4 ⋅ 6 ⋅ … ⋅ (2n)) / (2n)! ⋅ x^n using the Ratio Test. Learn about factorial simplification and series convergence.

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