To determine the radius of convergence ($R$) for the given series:

$\sum_{n=0}^\infty \frac{(n!)^2 x^{3n-1}}{(-4)^n (2n+3)!},$

we apply the ratio test for convergence.

Steps to Find the Radius of Convergence

1. The general term of the series is: $a_n = \frac{(n!)^2 x^{3n-1}}{(-4)^n (2n+3)!}.$

2. Using the ratio test, consider: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.$

   Substituting $a_n$ and $a_{n+1}$: $\frac{a_{n+1}}{a_n} = \frac{\frac{((n+1)!)^2 x^{3(n+1)-1}}{(-4)^{n+1} (2(n+1)+3)!}}{\frac{(n!)^2 x^{3n-1}}{(-4)^n (2n+3)!}}.$

3. Simplify the ratio: $\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2 x^{3n+2} \cdot (-4)^n \cdot (2n+3)!}{(n!)^2 x^{3n-1} \cdot (-4)^{n+1} \cdot (2(n+1)+3)!}.$

   • Use $(n+1)! = (n+1) \cdot n!$,
   • Simplify factorials and powers of $x$ and $(-4)$.

4. The key term simplifies to: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^2 \cdot x^3}{4 \cdot (2n+5)(2n+4)}.$

5. As $n \to \infty$: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x|^3}{4}.$

   For the series to converge, the ratio must satisfy: $\frac{|x|^3}{4} < 1.$

6. Solve for $|x|$: $|x|^3 < 4 \quad \implies \quad |x| < \sqrt[3]{4}.$

Radius of Convergence

The radius of convergence is: $R = \sqrt[3]{4}.$

Let me know if you want further clarifications or details!

5 Related Questions

1. How does the ratio test determine the convergence of a power series?
2. What is the significance of the radius of convergence in analyzing a series?
3. How do factorials influence the convergence of a series?
4. Could we use the root test instead of the ratio test for this series?
5. How do we handle convergence at the boundary points ($|x| = R$)?

Tip

When dealing with factorials and series, simplify as much as possible before applying limits—it reduces complexity and prevents errors.