Problem 33 - Thomas Calculus

Question

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=1}^{\infty} \frac{1}{2 \cdot 4 \cdot 6 \cdots(2 n)} x^{n} $$

Step-by-Step Solution

Answer

The radius of convergence is infinite, and the series converges absolutely for all \( x \). There are no values for conditional convergence.

Step 1: Express the General Term

First, express the given series in a simpler form. The general term of the series is \( \frac{x^n}{2 \cdot 4 \cdot 6 \cdots (2n)} \). Notice that the denominator is a product of even numbers, which can be expressed using the factorial notation as \( 2^n \cdot n! \). Thus, the general term becomes \( \frac{x^n}{2^n \cdot n!} \).

Step 2: Apply the Ratio Test for Convergence

To find the radius of convergence, use the ratio test. Consider the ratio of successive terms: \( \frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{2^{n+1} \cdot (n+1)!} \cdot \frac{2^n \cdot n!}{x^n} = \frac{x}{2(n+1)} \). Simplifying this ratio gives \( \left|\frac{x}{2(n+1)}\right| \), and applying the limit as \( n \to \infty \), we get \( \lim_{n \to \infty} \left|\frac{x}{2(n+1)}\right| = 0 \). This indicates that the series converges for all \( x \).

Step 3: Determine the Radius and Interval of Convergence

From Step 2, since the limit \( \lim_{n \to \infty} \left|\frac{x}{2(n+1)}\right| = 0 \) for all \( x \), the radius of convergence is \( R = \infty \) and the interval of convergence is \((-\infty, \infty)\).

Step 4: Identify Absolute Convergence

Since the series converges for all \( x \), it also converges absolutely for all \( x \). Absolute convergence implies that the series \( \sum \left|a_n\right| \) converges, which is satisfied for all \( x \) because we already established that it converges for all \( x \).

Step 5: Discuss Conditional Convergence

Conditional convergence happens when a series converges, but its absolute value does not converge. Here, since the series converges absolutely for all \( x \), there are no \( x \) values where the series converges conditionally.

Key Concepts

Absolute Convergence

Absolute convergence occurs when a series converges even if we take the absolute value of each of its terms. In simpler terms, if the series remains convergent when all terms are made positive, the series is said to be absolutely convergent.

For example, in the given series \[ \sum_{n=1}^{\infty} \frac{x^n}{2^n \cdot n!}\] we need to consider \(\sum \left| a_n \right|\).This results in the same series since all terms are already positive.

We already found the series converges for all values of \(x\). Since \(\sum \left| a_n \right|\) also converges for all \(x\), it means the original series converges absolutely everywhere. Absolute convergence is a strong form of convergence since it enforces convergence even when we manipulate the signs of the series' terms.

Conditional Convergence

Conditional convergence is a type of convergence that can occur when a series converges but does not converge absolutely. This might happen in series with both positive and negative terms, leading to the balance needed for convergence without the absolute condition.

However, in the series at hand, \( \sum_{n=1}^{\infty} \frac{x^n}{2^n \cdot n!} \), the series converges absolutely for all values of \(x\) as discussed.

That means there is no value of \(x\) for which the series might converge conditionally. In general, conditional convergence is less common than absolute convergence and requires careful analysis of the sum without taking absolute values.

Ratio Test

The ratio test is a powerful tool used to determine the convergence or divergence of a series, particularly useful for series with factorials or exponential terms. To apply the ratio test, we examine the limit of the absolute value of the ratio of successive terms.

For the series \( \sum_{n=1}^{\infty} \frac{x^n}{2^n n!} \), calculate the ratio \( \frac{a_{n+1}}{a_n} = \left| \frac{x^{n+1}}{2^{n+1} (n+1)!} \cdot \frac{2^n n!}{x^n} \right| \).
Simplifying gives \( \left| \frac{x}{2(n+1)} \right| \).
By taking the limit as \( n \to \infty \), we find \( \lim_{n \to \infty} \left| \frac{x}{2(n+1)} \right| = 0 \), indicating convergence.

If the limit of the ratio is less than 1, the series converges absolutely. Since our example yields 0, it converges for all \(x\), and thus the radius of convergence is \(R = \infty\). The ratio test confirmed the series converges everywhere, reflecting strong absolute convergence.