Problem 37 - Thomas Calculus

Question

In Exercises \(37-40,\) find the series' radius of convergence. $$ \sum_{n=1}^{\infty} \frac{n !}{3 \cdot 6 \cdot 9 \cdot \cdot \cdot 3 n} x^{n} $$

Step-by-Step Solution

Answer

The radius of convergence is 3.

Step 1: Understand the Series

The given series is \( \sum_{n=1}^{\infty} \frac{n !}{3 \cdot 6 \cdot 9 \cdots 3n} x^{n} \). This is an infinite series where each term is described as \( \frac{n!}{3^n \cdot n!} x^n \) since the denominator can be expressed as \( (3^n \cdot n!) \). This simplifies to \( \frac{x^n}{3^n} \).

Step 2: Apply the Ratio Test

To determine the radius of convergence, apply the Ratio Test. According to this test, consider \( a_n = \frac{x^n}{3^n} \). Calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

Step 3: Simplify the Ratio

Now, calculate \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{x^{n+1}}{3^{n+1}}}{\frac{x^n}{3^n}} \right| = \left| \frac{x \cdot 3^n}{3^{n+1}} \right| = \left| \frac{x}{3} \right| \).

Step 4: Determine Convergence Condition

For convergence, the Ratio Test requires \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). Therefore, \( \left| \frac{x}{3} \right| < 1 \) implies \( |x| < 3 \).

Step 5: Identify the Radius of Convergence

The inequality \( |x| < 3 \) means that the radius of convergence \( R = 3 \).

Key Concepts

Ratio Test

The ratio test is a method used to determine the convergence or divergence of an infinite series. It's an effective technique especially when the series terms involve factorials or exponential functions. The test requires you to examine the limit of the ratio of consecutive terms \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If this limit is less than 1, the series converges. Conversely, if the limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive, and another method must be used. For the given series, the ratio test revealed that \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x}{3} \right|\]
which led to the condition \( |x| < 3 \). This indicates the radius of convergence is 3, and the series converges for values of \( x \) within this range.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. In mathematical terms, it can be written as \( \sum_{n=1}^{\infty} a_n \). Infinite series can either converge, meaning they approach a specific value, or diverge, where they do not settle to a finite number. To determine if an infinite series converges, various tests like the ratio test, root test, and integral test can be utilized, depending on the series' form. For the problem in question, the infinite series \( \sum_{n=1}^{\infty} \frac{n!}{3 \cdot 6 \cdot 9 \cdots 3n} x^n \) was analyzed for convergence within a certain radius. By simplifying and applying the ratio test, it was established that the series converges when \( |x| < 3 \). Understanding the nature of infinite series is crucial in calculus as they frequently appear in mathematical modeling and analysis.

Factorial

Factorial, denoted by \( n! \), is the product of all positive integers up to \( n \). It is a fundamental concept in mathematics, particularly in combinatorics, algebra, and calculus.The formula for a factorial is given by:

• \( 0! = 1 \) (by definition)
• \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \) for \( n \geq 1 \)

Factorials grow very quickly as \( n \) increases due to their multiplicative nature. This concept was crucial in simplifying the given series in the exercise. By recognizing that the original expression \( \frac{n!}{3 \cdot 6 \cdot 9 \cdots 3n} \) equates to a more manageable form involving division by \( (3^n \cdot n!) \), the students can focus on assessing convergence effectively.