Problem 35 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{(n+3) !}{3 ! n ! 3^{n}} $$

Step-by-Step Solution

Answer

The series converges by the ratio test.

Step 1: Rewrite the Series

The series given is \( \sum_{n=1}^{\infty} \frac{(n+3)!}{3!n!3^n} \). Our goal is to determine its convergence. Let's begin by rewriting the general term of the series so it's easier to manage. We already have the term \( \frac{(n+3)!}{3!n!3^n} \).

Step 2: Simplify the General Term

Using the factorial property \( (n+3)! = (n+3)(n+2)(n+1)n! \), the general term can be expressed as:\[\frac{(n+3)(n+2)(n+1)n!}{3!n!3^n} = \frac{(n+3)(n+2)(n+1)}{6 \cdot 3^n}\].

Step 3: Check for Convergence using the Ratio Test

We apply the ratio test, which uses the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series converges absolutely. Let \( a_n = \frac{(n+3)(n+2)(n+1)}{6 \cdot 3^n} \) and find \( a_{n+1} = \frac{(n+4)(n+3)(n+2)}{6 \cdot 3^{n+1}} \).

Step 4: Calculate the Ratio

Calculate the ratio:\[\frac{a_{n+1}}{a_n} = \frac{(n+4)(n+3)(n+2) \cdot 3^n}{(n+3)(n+2)(n+1) \cdot 3^{n+1}} = \frac{n+4}{3(n+1)}\].

Step 5: Take the Limit of the Ratio

Evaluate the limit:\[\lim_{n \to \infty} \frac{n+4}{3(n+1)} = \lim_{n \to \infty} \frac{1 + \frac{4}{n}}{3 + \frac{3}{n}} = \frac{1}{3}\].Since \( \frac{1}{3} < 1 \), by the ratio test, the series converges.

Key Concepts

Factorial Simplification

In the given series, we encounter a factorial expression \( (n+3)! \), which can be cumbersome to handle directly. Factorials are products of consecutive integers from a number down to 1, like \( n! = n \times (n-1) \times ... \times 1 \). Here, \( (n+3)! \) means \( (n+3) \times (n+2) \times (n+1) \times n! \). This allows for simplification in expressions like \( \frac{(n+3)!}{n!} \) because most terms cancel out, leaving you with a more manageable expression of \( (n+3)(n+2)(n+1) \). \
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By simplifying factorials, we can significantly reduce complexity, both in writing and calculations. It's analogous to simplifying fractions where common terms cancel out. This step is crucial in preparing the series for further analysis, like applying tests for convergence.

Ratio Test

The ratio test is an indispensable tool when dealing with series, particularly those involving factorials and exponential terms. The main idea is to determine how each term in a series compares to the next as \( n \) tends to infinity. The test provides a simple criterion: consider \( a_n \) as a general term of the series; compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). \
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If this limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. If it equals 1, the test is inconclusive. \
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For the series given, identifying the \( a_n \) term, carefully calculating \( a_{n+1} \), and simplifying the ratio \( \frac{a_{n+1}}{a_n} \) led us to deduce convergence, since the limit was calculated to be \( \frac{1}{3} \), which is decisively less than 1.

Infinite Series

An infinite series is the sum of an infinite sequence of terms. Symbolically, it's expressed as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) are the terms of the series. Series can converge (sum to a finite value) or diverge (grow without bound). Infinity in this context poses an interesting challenge: you can't literally "add all terms," so we rely on mathematical techniques to infer the series' behavior. \
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Convergent series do not necessarily sum to 0 but to a finite number, reflecting some equilibrium point. Divergent series, on the other hand, indicate growth trends or instability. \
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Various methods exist to test convergence, including the comparison test, integral test, and, as used here, the ratio test. Understanding series lays fundamental groundwork for calculus and is crucial in various applications across mathematics and physics.

Limit Evaluation

Evaluating limits is a foundational concept in calculus used to understand behavior as variables approach certain values, often infinity. In the context of series, limits help decide convergence. In the ratio test, the evaluation of \( \lim_{n \to \infty} \frac{n+4}{3(n+1)} \) was critical. \
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By evaluating \( \lim_{n \to \infty} \frac{1 + \frac{4}{n}}{3 + \frac{3}{n}} \), the presence of terms like \( \frac{4}{n} \) highlights how negligible smaller terms become as \( n \) grows. Essentially \( \frac{4}{n} \to 0 \), simplifying the expression to \( \frac{1}{3} \). \
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Understanding limits involves intuition about growth, decay, and balance in expressions, showing how "big" parts of expressions behave relative to "small" parts as certain variables increase or decrease indefinitely.