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The given series is:

\[
\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}
\]

To find the radius of convergence \( R \) and the interval of convergence, we can use the **Ratio Test**. The Ratio Test states that for a series \(\sum a_n\), the radius of convergence \(R\) can be found by examining the limit:

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

### Step 1: Identify \( a_n \)
In this series, we have:
\[
a_n = \frac{(-1)^n x^{2n}}{(2n)!}
\]

### Step 2: Apply the Ratio Test
Using the Ratio Test, we calculate:

\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{(2(n+1))!}}{\frac{(-1)^n x^{2n}}{(2n)!}} \right|
\]

Simplify this expression:

\[
= \left| \frac{(-1)^{n+1} x^{2n+2}}{(2n+2)!} \cdot \frac{(2n)!}{(-1)^n x^{2n}} \right|
= \left| \frac{x^2}{(2n+2)(2n+1)} \right|
\]

As \(n \to \infty\), the terms \((2n+2)(2n+1)\) grow large, causing the fraction to approach zero:

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0
\]

### Step 3: Interpret the Result
Since this limit is zero (which is less than 1) for all values of \(x\), the series converges for all \(x\) values.

### Conclusion
The radius of convergence \( R \) is infinite, meaning the series converges for all real \(x\). Therefore, the **interval of convergence** is:

\[
(-\infty, \infty)
\]

Would you like a further breakdown or have any specific questions?

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**Related Questions:**

1. What would happen to the radius of convergence if we used a different test?
2. How does the factorial in the denominator influence the convergence of the series?
3. What is the significance of the alternating factor \((-1)^n\) in this series?
4. How would the series change if we had \( (2n+1)! \) in the denominator instead?
5. What if \(x\) were replaced by a complex variable? How would convergence be affected?

**Tip:** The Ratio Test is particularly useful for series with factorials or exponential terms due to its reliance on the limit of the ratio between terms.

Find the radius of convergence and interval of convergence of the given series using the ratio test.

\[ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \]

Find the radius and interval of convergence for the series \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \) using the Ratio Test. Detailed step-by-step solution included. Suitable for advanced high school and college-level calculus.

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