Does This Series Converge? A Detailed Analysis

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of infinite series. Specifically, we'll be tackling the question: Does the following series converge or diverge? We'll explore the series step-by-step, providing detailed reasoning for our answer. So, buckle up, grab your coffee, and let's get started!

$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot (2 n-1)}{8^n 2^n n!}$

Understanding the Problem: Convergence vs. Divergence

Before we begin, let's refresh our memory on the core concepts. When we talk about an infinite series, we're essentially adding up an infinite number of terms. The big question is: does this sum approach a finite value (converge), or does it grow without bound (diverge)? Understanding this is critical in many areas of mathematics, physics, engineering, and computer science. Convergence means the series has a finite sum, and divergence means the sum does not exist or goes to infinity. We need to determine the behavior of the given series.

Our series is a bit complex, but don't worry, we'll break it down piece by piece. The numerator involves a product of odd numbers, the denominator has powers of 8 and 2, and we also have the familiar factorial function in play. This blend suggests that the ratio test could be a good approach to investigate its convergence. We will now go through this step by step. We have to analyze the limit and determine whether it converges or diverges.

Now, let's look at the series closely. This series involves a product of odd numbers in the numerator, powers of 8 and 2 in the denominator, and a factorial in the denominator as well. This setup is a classic indicator that the ratio test might be the right tool for the job. Our goal is to calculate the limit and make a determination whether it converges or diverges.

Applying the Ratio Test: The Heart of the Matter

The Ratio Test is a powerful tool for determining the convergence or divergence of a series. The ratio test is useful for series involving factorials and exponential terms because these elements make the ratio easier to evaluate. It works by examining the ratio of consecutive terms in the series. Let's denote the nth term of our series as $a_n$. So,

$a_n = \frac{1 \cdot 3 \cdot \cdots \cdot (2 n-1)}{8^n 2^n n!}$

The first step in the ratio test is to compute the ratio of consecutive terms: $\frac{a_{n+1}}{a_n}$. This is the core of our analysis. We are using the ratio test here because it provides a convenient method for analyzing the behavior of series with factorials and other terms. The ratio test allows us to compare successive terms, which will help us determine if the series converges or diverges. The limit of this ratio gives us the required information.

So, let's calculate $a_{n+1}$:

$a_{n+1} = \frac{1 \cdot 3 \cdot \cdots \cdot (2n-1) \cdot (2(n+1)-1)}{8^{n+1} 2^{n+1} (n+1)!}$

Simplifying, we get:

$a_{n+1} = \frac{1 \cdot 3 \cdot \cdots \cdot (2n-1) \cdot (2n+1)}{8^{n+1} 2^{n+1} (n+1)!}$

Now, let's calculate the ratio $\frac{a_{n+1}}{a_n}$:

$\frac{a_{n+1}}{a_n} = \frac{\frac{1 \cdot 3 \cdot \cdots \cdot (2n-1) \cdot (2n+1)}{8^{n+1} 2^{n+1} (n+1)!}}{\frac{1 \cdot 3 \cdot \cdots \cdot (2 n-1)}{8^n 2^n n!}}$

Simplifying this complex fraction gives us:

$\frac{a_{n+1}}{a_n} = \frac{1 \cdot 3 \cdot \cdots \cdot (2n-1) \cdot (2n+1)}{8^{n+1} 2^{n+1} (n+1)!} \cdot \frac{8^n 2^n n!}{1 \cdot 3 \cdot \cdots \cdot (2 n-1)}$

Which simplifies to:

$\frac{a_{n+1}}{a_n} = \frac{2n+1}{8 \cdot 2 \cdot (n+1)} = \frac{2n+1}{16(n+1)}$

Now, to determine the convergence or divergence, we need to find the limit of this ratio as n approaches infinity:

$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{2n+1}{16(n+1)}$

To evaluate this limit, we can divide both the numerator and denominator by n:

$\lim_{n\to\infty} \frac{2n+1}{16(n+1)} = \lim_{n\to\infty} \frac{2 + \frac{1}{n}}{16(1 + \frac{1}{n})}$

As n approaches infinity, $\frac{1}{n}$ approaches 0. Therefore, the limit becomes:

$\lim_{n\to\infty} \frac{2 + \frac{1}{n}}{16(1 + \frac{1}{n})} = \frac{2}{16} = \frac{1}{8}$

The ratio test states that if the limit of the ratio is less than 1, the series converges. If it is greater than 1, the series diverges. If the limit is equal to 1, the test is inconclusive. In our case, the limit is $\frac{1}{8}$, which is less than 1. Therefore, our series converges.

Conclusion: Convergence Confirmed!

Based on the Ratio Test analysis, we've determined that the series $\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot (2 n-1)}{8^n 2^n n!}$ converges. This means the sum of the infinite terms approaches a finite value. This insight is extremely useful in mathematics. The series converges to a certain value. Using the ratio test, we've seen it converges because the limit of the ratio is less than 1.

Our journey through this series reveals the power of the Ratio Test. As we have seen, the test is not only applicable to series involving factorials but is also very effective with other complex types of series. The Ratio Test provides a powerful method for determining the convergence of these series. The result is the series converges. Always remember to check the value of the limit to ensure it is less than 1, which will prove the series' convergence.

I hope you enjoyed this deep dive into series convergence. Keep practicing, and you'll become a master of infinite series in no time. If you have any questions, feel free to ask! Happy calculating!

Final Answer and Explanation

• The series converges. We determined this using the Ratio Test.
• The limit of the ratio $\frac{a_{n+1}}{a_n}$ is $\frac{1}{8}$, which is less than 1.
• A limit less than 1 implies convergence.

In summary, the series converges because the limit obtained from the Ratio Test is less than 1.

For further reading and more examples, I recommend checking out these resources:

• Khan Academy - Ratio Test – Khan Academy offers excellent video explanations and practice problems.