Question

Determine the radius of convergence of the given power series. $$ \sum_{n=0}^{\infty} \frac{x^{2 n}}{n !} $$

Short answer

Answer: The radius of convergence of the given power series is ∞.

Step-by-step solution

Apply the Ratio Test

Applying the ratio test, we need to find the limit: $$ \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{\frac{x^{2(n+1)}}{(n+1)!}}{\frac{x^{2n}}{n!}} $$

Simplify the Ratio

Let's simplify the ratio before finding the limit: $$ \lim_{n\to\infty} \frac{x^{2(n+1)}}{(n+1)!} \cdot \frac{n!}{x^{2n}} = \lim_{n\to\infty} \frac{x^{2n+2}}{(n+1)!} \cdot \frac{n!}{x^{2n}} $$ Now, we can cancel \(x^{2n}\) and \(n!\) in the numerator and denominator: $$ \lim_{n\to\infty} \frac{x^2}{(n+1)!} \cdot \frac{n!}{1} = \lim_{n\to\infty} \frac{x^2 n!}{(n+1)!} $$

Compute the Limit

Next, we compute the limit as \(n \to \infty\): $$ \lim_{n\to\infty} \frac{x^2 n!}{(n+1)!} = \lim_{n\to\infty} \frac{x^2 n!}{(n+1)n!} $$ We can cancel out \(n!\) in the numerator and denominator: $$ \lim_{n\to\infty} \frac{x^2}{(n+1)} = 0 $$ As the limit is zero for any value of x, the ratio test always gives a value less than 1, which means the power series converges for all x.

Conclusion

The radius of convergence is given as: $$ R = \lim_{n\to\infty} \frac{1}{\lim_{n\to\infty} \frac{x^2}{(n+1)}} = \lim_{n\to\infty} (n+1) = \infty $$ Thus, the radius of convergence of the given power series is \(\infty\), and the series converges for all values of x.