Problem 28 - Calculus: Early Transcendentals

Question

Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {n!x^n}{1 \cdot 3 \cdot 5 \cdot \cdot \cdot \cdot \cdot (2n - 1)} \)

Step-by-Step Solution

Answer

Radius of convergence is 2; interval of convergence is \((-2, 2)\).

Step 1: Understanding the General Term

The general term of the series is given as \( a_n = \frac{n! \, x^n}{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} \). This part is crucial as it forms the basis for applying the ratio test to find the radius of convergence.

Step 2: Applying the Ratio Test

To determine the radius of convergence, apply the ratio test: \( \lim_{n \to \infty} \left| \frac{ a_{n+1} }{ a_n } \right| \). Substituting the general term, we have \[ \lim_{n \to \infty} \left| \frac{ (n+1)! \, x^{n+1} }{1 \cdot 3 \cdot \cdots \cdot (2n+1)} \times \frac{1 \cdot 3 \cdot \cdots \cdot (2n-1)}{n! \, x^n} \right|. \] Simplifying gives: \[ \lim_{n \to \infty} \left| x \right| \times \frac{n+1}{2n+1}. \]

Step 3: Evaluating the Limit

Further simplify the expression obtained from the ratio test: \[ \lim_{n \to \infty} \left| x \right| \times \frac{n+1}{2n+1} = \left| x \right| \lim_{n \to \infty} \frac{n+1}{2n+1}. \] As \( n \to \infty \), this fraction approaches \( \frac{1}{2} \), thus we have \( \frac{1}{2} |x| \).

Step 4: Solve for Radius of Convergence

For convergence, \( \frac{1}{2} |x| < 1 \). Solving this inequality gives \( |x| < 2 \). Therefore, the radius of convergence \( R \) is 2.

Step 5: Determine Interval of Convergence

Start with the inequality \( |x| < 2 \) giving the interval \( -2 < x < 2 \). Then check the endpoints \( x = -2 \) and \( x = 2 \): - When \( x = 2 \), the series becomes \( \sum_{n=1}^{\infty} \frac{n! \, 2^n}{1 \cdot 3 \cdot \cdots \cdot (2n-1)} \), which diverges due to the factorial in the numerator dominating.- When \( x = -2 \), the series becomes \( \sum_{n=1}^{\infty} \frac{n! \, (-2)^n}{1 \cdot 3 \cdot \cdots \cdot (2n-1)} \), which also diverges for the same reason.Therefore, the interval of convergence is \((-2, 2) \).

Key Concepts

Interval of Convergence

The interval of convergence defines the range of values for which a series converges. For any power series like \( \sum a_n x^n \), once you determine the radius of convergence, you typically find the interval of convergence by testing or analyzing endpoint behavior. In our example problem, after calculating the radius of convergence, the next task is to test whether the series converges at each endpoint of the interval.
For a radius of 2 with the interval from \(-2\) to \(2\), testing the endpoints \(x = 2\) and \(x = -2\) helps to confirm that these are not included because the series diverges at these points. Thus, the interval of convergence is open, denoted as \((-2, 2)\).
Through this, we ensure that above all else, each value of \(x\) within the interval results in a convergent series.

Ratio Test

The ratio test is an essential tool when dealing with series to determine convergence. It involves taking the limit of the ratio of successive terms of the series. For the series given in our example:

• The general term is \( a_n = \frac{n! \cdot x^n}{1 \cdot 3 \cdot 5 \cdots (2n-1)} \).
• Applying the ratio test, calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

Substitute the general term into this limit expression and simplify. If the limit is less than one, the series converges absolutely within that region of \(x\) values. In this problem, the limit simplifies to \( \lim_{n \to \infty} \left| x \right| \cdot \frac{n+1}{2n+1} \) which eventually leads to \( \frac{1}{2} |x| \).
For convergence, this result tells us \( \frac{1}{2} |x| < 1 \). Solving for \(|x|\), we find \(|x| < 2\), which gives us the radius of convergence as 2.

Factorial

Factorials, denoted by the exclamation mark \(!\), are mathematical expressions where \(n!\) ("n factorial") is the product of all positive integers less than or equal to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow rapidly with increasing \(n\).
In our series example, we see \(n!\) in the numerator of the general term, \(\frac{n! \cdot x^n}{1 \cdot 3 \cdot 5 \cdots (2n-1)}\).
The presence of factorials is significant, often causing divergence when not properly balanced by the rest of the series' terms, as demonstrated by the failure of convergence at the endpoints \(x = 2\) and \(x = -2\). Due to the factorial's size, it overpowers the corresponding odd-product denominator, leading to divergence.

Series Convergence

Series convergence refers to the condition under which a given infinite series approaches a finite limit. Convergence of a series is fundamental to understanding many concepts in calculus and analysis.
In the context of this problem, convergence is examined primarily through the application of the ratio test. Given the steps shown earlier, we found that the series converges for values of \(|x|\) less than 2; hence the radius of convergence is 2.
Aside from this test, different techniques may be used to check for convergence or divergence at specific points, particularly at interval boundaries, ensuring a comprehensive understanding of the series' behavior. Overall, knowing whether a series converges helps predict the series' utility and reliability for different mathematics and science applications.