Math Problem Statement

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

Solution

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

∑n=1∞2⋅4⋅6⋅⋯⋅(2n)(2n)!xn\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⋅4⋅6⋅⋯⋅(2n)2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n) can be rewritten as 2n⋅n!2^n \cdot n! since it is the product of the first nn even numbers. Substituting this into the series, the nn-th term becomes:

an=2n⋅n!(2n)!.a_n = \frac{2^n \cdot n!}{(2n)!}.

Thus, the series becomes:

∑n=1∞2n⋅n!(2n)!xn.\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→∞∣an+1an∣<1.\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.

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

Dividing an+1a_{n+1} by ana_n:

= \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.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Radius of Convergence
Ratio Test
Factorial

Formulas

a_n = (2^n ⋅ n!) / (2n)!
Ratio Test: lim (n → ∞) |a_{n+1} / a_n| < 1

Theorems

Ratio Test for Convergence

Suitable Grade Level

Undergraduate Mathematics

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