Problem 7 - Thomas Calculus

Question

In Exercises \(1-8,\) use the Ratio Test to determine if each series conyerges ahsolutely or diveroes. $$\sum_{n=1}^{\infty}(-1)^{n} \frac{n^{2}(n+2) !}{n ! 3^{2 n}}$$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Write Down the General Term

The general term of the series given is \( a_n = (-1)^n \frac{n^2 (n+2)!}{n! 3^{2n}} \). The absolute value for the ratio test is considered, so \(|a_n| = \frac{n^2 (n+2)!}{n! (3^{2n})} \).

Step 2: Set Up the Ratio \( \frac{a_{n+1}}{a_n} \)

To apply the Ratio Test, calculate the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \). This becomes: \[\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^2 ((n+3)!) \cdot n! \cdot 3^{2n}}{n^2 (n+2)! (n+1)! \cdot 3^{2(n+1)}}.\]

Step 3: Simplify the Expression

Simplify the fraction:- The \( (n+3)! \) in the numerator and the \( (n+2)! \) in the denominator can be reduced to \( (n+3) \).- \( 3^{2(n+1)} = 9 \cdot 3^{2n} \). So, it simplifies to:\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^2 (n+3) \cdot 3^{2n}}{n^2 \cdot 9 \cdot 3^{2n}} = \frac{(n+1)^2 (n+3)}{9n^2}.\]

Step 4: Take the Limit as \( n \to \infty \)

Now, take the limit of the simplified expression:\[\lim_{n \to \infty} \frac{(n+1)^2 (n+3)}{9n^2}.\]

Step 5: Evaluate the Limit

The highest power of \( n \) in the numerator is \( n^3 \), and in the denominator, it's also \( n^3 \). Thus, the limit evaluates to:\[\lim_{n \to \infty} \frac{n^3}{9n^2} = \lim_{n \to \infty} \frac{n^3}{9n^3} = \frac{1}{9}.\]

Step 6: Conclusion Using the Ratio Test

The limit \( \frac{1}{9} \) is less than 1. According to the Ratio Test, since this limit is less than 1, the series converges absolutely.

Key Concepts

Series Convergence

Understanding whether a series converges or diverges is crucial in mathematics as it helps identify the behavior of series over an infinite interval. A series is an expression with an infinite sum of terms. To determine if it converges, we confirm whether its terms approach a finite value. For example, the series \[ \sum_{n=1}^{\infty} (-1)^n \frac{n^2 (n+2)!}{n! 3^{2n}} \] requires evaluation to see if this sum converges to a specific number as the number of terms grows infinitely large. Generally, the series convergence is identified through various tests, including the Ratio Test, which comes in handy when we deal with factorials or terms growing rapidly. Recognizing convergence not only ensures understanding of the series behavior but significantly aids in applications such as physics and engineering, where series represent time intervals or motions over infinity.

Factorials

Factorials play a significant role in series and sequences, especially when evaluating convergence using tests like the Ratio Test. A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). It often appears in series to express growth or patterns effectively.

When simplifying factorial expressions, recognizing successive terms is vital. For instance, \((n+2)!\) can be simplified against \((n+1)!\) or \((n)!\) by expanding the sequence. In the given series,

• The term \((n+2)!\) was simplified over \((n+1)!\), resulting in just additional multiplicands like \((n+2)\), \((n+1)\).
• This simplification is key when reducing complex factorial ratios to more manageable forms for further limit evaluation.

Limit Evaluation

Limit evaluation is an essential part of the Ratio Test and in determining series convergence. By simplifying expressions into manageable fractions, limits help us predict the behavior of series as \( n \) approaches infinity. In the example provided, after simplifying the ratio \( \frac{a_{n+1}}{a_n} \), the process involves checking:

• The degrees of growth in both the numerator \((n+1)^2 (n+3)\) and denominator \(9n^2\).
• The highest power of \( n \) in both becomes crucial. With both having the highest power \(n^3\), dividing both gives insight into the convergence state.

The limit \[ \lim_{n \to \infty} \frac{(n+1)^2 (n+3)}{9n^2} = \frac{1}{9} \] is key in this context. Since this result is less than 1, it signals convergence under the Ratio Test. Thus, seeing how easily the behavior of complex terms boils down to a simple limit can be truly insightful in convergence analysis.