Problem 33 - Single Variable Calculus

Question

Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty} \frac{4 \cdot 7 \cdot 10 \cdot \cdots \cdot(3 n+1)}{4^{n}(n+1) !}\)

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{4 \cdot 7 \cdot 10 \cdot \cdots \cdot(3n+1)}{4^{n}(n+1) !} \) is divergent. This conclusion is reached by applying the Ratio Test, which gives a limit \( L = 3 \) (greater than 1).

Step 1: Write down the general terms of the series

The given series is \[ \sum_{n=1}^{\infty} \frac{4 \cdot 7 \cdot 10 \cdot \cdots \cdot(3n+1)}{4^{n}(n+1) !} \] Notice that the denominator contains the factorial of \( n+1 \) while the numerator contains a sequence of terms which are multiples of 3 added to 1.

Step 2: Apply the Ratio Test

To find the limit, we consider the ratio \[ L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_{n}} \right| \] where \( a_{n} = \frac{4 \cdot 7 \cdot 10 \cdot \cdots \cdot(3n+1)}{4^{n}(n+1) !} \) and \( a_{n+1} = \frac{4 \cdot 7 \cdot 10 \cdot \cdots \cdot(3(n+1)+1)}{4^{n+1}((n+1) +1) !} \). By simplifying, \[ L = \lim_{{n \to \infty}} \frac{(3n+4)}{(n+2)} \]

Step 3: Compute the limit

As \( n \) approaches infinity, we find that \[ L = \lim_{{n \to \infty}} \frac{(3n+4)}{(n+2)} = 3 \]

Step 4: Determine convergence or divergence

Because \( L > 1 \), we conclude that the series is divergent according to the Ratio Test. As such, the series \( \sum_{n=1}^{\infty} \frac{4 \cdot 7 \cdot 10 \cdot \cdots \cdot(3n+1)}{4^{n}(n+1) !} \) is divergent.

Key Concepts

Ratio Test

The Ratio Test is a tool used to determine the convergence or divergence of infinite series, especially useful when handling factorials or exponential terms. In essence, the test evaluates the limit of the absolute value of the ratio between consecutive terms in the series. Here's how it works:

• You first need to identify the general term of the series. Let's call this term \( a_n \).
• Next, find the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \), which involves taking the ratio of the subsequent term \( a_{n+1} \) to the current term \( a_n \).
• Then, compute the limit \( L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \).

The results will guide you:

• If \( L < 1 \), the series is absolutely convergent.
• If \( L = 1 \), the ratio test is inconclusive, and other methods should be used.
• If \( L > 1 \), the series is divergent.

In this exercise, calculating \( L \) for our series resulted in 3, a clear sign that the series is divergent.

Divergent Series

When a series is said to be divergent, it means that as you add more and more terms, the sum doesn't settle toward a specific number. Instead, it tends to increase or oscillate indefinitely.
Divergence can be identified in a variety of ways:

• One common method is the Ratio Test, where if the limit \( L > 1 \), it confirms divergence.
• Another indication of divergence is the presence of terms that do not approach zero. For a series \( \sum a_n \) to converge, \( a_n \) must approach zero as \( n \to \infty \).

In our example, the ratio test clearly showed a divergence since the limit \( L \) was equal to 3, which is greater than 1. This means that the terms in the series do not shrink quickly enough to converge to a particular value.

Factorial Terms

Factorial terms, denoted as \( n! \), represent the product of all integers from 1 up to \( n \). These terms play a crucial role in many series, especially those involving rapid growth.In the given series, the denominator includes \( (n+1)! \), which grows significantly fast as \( n \) increases:

• Facilitates simplification when combined with the numerator, often involving exponential growth terms such as \( 3n+1 \).
• Counterbalances growth in the numerator, determining the overall behavior of the series as it progresses toward infinity.

Factorials are key in mathematical series problems because their rapid growth can either converge, depending on interaction with other factors, or can significantly impact tests like the Ratio Test. In our exercise, despite the factorial in the denominator, the terms in the numerator grew faster, resulting in the series being divergent.