Problem 62 - Thomas Calculus

Question

Convergence or Divergence Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)}$$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Write the General Term

The given series is \( \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)} \). The general term \( a_n \) of the series can be expressed as \( a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)(3^n + 1)} \).

Step 2: Simplify the General Term Using Double Factorials

Note that the numerator appears to be the product of the first \( n \) odd numbers, which can be expressed as \( (2n-1)!! \). Similarly, the denominator is the product of the first \( n \) even numbers, or \( (2^n \cdot n!) \). Therefore, \( a_n = \frac{(2n-1)!!}{2^n n! (3^n + 1)} \).

Step 3: Apply Ratio Test for Convergence

To determine the convergence of the series, we use the Ratio Test. Consider: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(2n+1)!!}{2^{n+1} (n+1)!(3^{n+1}+1)} \times \frac{2^n n!(3^n+1)}{(2n-1)!!} \right| \).

Step 4: Evaluate the Limit

Simplifying the expression from Step 3 yields:\[\lim_{n \to \infty} \frac{(2n+1)(3^{n}+1)}{2(3^{n+1}+1)} \cdot \frac{1}{n+1} = \lim_{n \to \infty} \frac{(2n+1)(3^n+1)}{2 \cdot 3 \cdot 3^n} \cdot \frac{1}{n+1}\] After canceling terms and simplifying further, we focus primarily on:\[\lim_{n \to \infty} \frac{2n+1}{6n+6}\] Which simplifies and approaches \(\frac{1}{3}\) as \(n\) goes to infinity.

Step 5: Conclude Based on the Ratio Test

Since the limit obtained in Step 4 is \(\frac{1}{3}\), which is less than 1, according to the Ratio Test, the series converges. Thus, the series \( \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a method used to determine the convergence of an infinite series. It's especially useful when the series includes factorials or exponential terms. To apply the test, consider the series with terms \(a_n\). We compute the limit:

\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]

• If \(L < 1\), the series converges.
• If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive.

In the provided example, the limit was calculated to be \(\frac{1}{3}\). Since this is less than 1, the Ratio Test confirms that the series converges. This means that as you add more and more terms, the sum of the series approaches a specific finite value.
The Ratio Test is powerful but sometimes might not give a definitive answer if it ends up being equal to 1.

Infinite Series

An infinite series is the sum of the terms of an infinitive sequence. It is generally written as:

\[\sum_{n=1}^{\infty} a_n\]Each term \(a_n\) of the sequence contributes to the total sum. Even though infinite series can seem daunting, many of them do converge.These series play a critical role in calculus and analysis, allowing mathematicians to solve problems that involve endless processes.
In the context of the problem, the given series, \(\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)}\), is an infinite series. Using the Ratio Test, we found that it converges, indicating its sum approaches a finite value as more terms are added.
Infinite series can often represent functions, probabilities, or other mathematical expressions in a compact form.

Factorial

Factorial is an important mathematical concept symbolized by an exclamation mark \(!\). The factorial of a non-negative integer \(n\) is defined as the product of all positive integers less than or equal to \(n\). It's expressed as:

\[n! = n \times (n-1) \times (n-2) \times \cdots \times 1\]Factorials grow very quickly, which is why they're common in problems dealing with large calculations.In the exercise's series, double factorials are also used, which apply to odd numbers in the numerator and even numbers in the denominator. The double factorial \((2n-1)!!\) means multiplying all the odd numbers up to \(2n-1\), whereas \((2n)!!\) involves the product of all even numbers up to \(2n\).
Understanding factorials helps simplify complex terms and supports convergence tests. Factorials often appear in Taylor series, combinatorics, and probability.