Problem 36 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{n 2^{n}(n+1) !}{3^{n} n !} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Simplify the Expression

The given series is \( \sum_{n=1}^{\infty} \frac{n \cdot 2^{n} (n+1) !}{3^{n} \cdot n !} \). Simplify the term \( \frac{(n+1)!}{n!} \) as it equals \( n + 1 \). The series then becomes:\[ \sum_{n=1}^{\infty} \frac{n \cdot 2^n (n+1)}{3^n} \] which further simplifies to: \[ \sum_{n=1}^{\infty} \frac{n(n+1) \cdot 2^n}{3^n} \].

Step 2: Apply Ratio Test

The Ratio Test involves taking the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as \( n \to \infty \). Let \( a_n = \frac{n(n+1) \cdot 2^n}{3^n} \). Then \[ a_{n+1} = \frac{(n+1)(n+2) \cdot 2^{n+1}}{3^{n+1}} \]. Compute \( \left| \frac{a_{n+1}}{a_n} \right| \):\[ \left| \frac{(n+1)(n+2) \cdot 2^{n+1} \cdot 3^n}{n(n+1) \cdot 2^n \cdot 3^{n+1}} \right| = \left| \frac{(n+2) \cdot 2}{n \cdot 3} \right| = \frac{2(n+2)}{3n} \].

Step 3: Evaluate the Limit

Evaluate the limit as \( n \to \infty \):\[ \lim_{n \to \infty} \frac{2(n+2)}{3n} = \lim_{n \to \infty} \frac{2n+4}{3n} = \lim_{n \to \infty} \frac{2 + \frac{4}{n}}{3} = \frac{2}{3} \].

Step 4: Conclude Convergence or Divergence

Since the Ratio Test limit \( \frac{2}{3} \) is less than 1, the given series \( \sum_{n=1}^{\infty} \frac{n(n+1) \cdot 2^n}{3^n} \) converges by the Ratio Test.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence of infinite series. It's particularly useful for series where terms contain factorials, exponents, and products. To use the Ratio Test:

• Take the absolute value of the ratio of successive terms in the series: \( \left| \frac{a_{n+1}}{a_n} \right| \)
• Evaluate the limit of this ratio as \( n \to \infty \)

If the limit is:

• Less than 1, the series converges.
• Greater than 1, the series diverges.
• Equal to 1, the test is inconclusive.

In our example, the ratio simplifies to \( \frac{2(n+2)}{3n} \), which has a limit of \( \frac{2}{3} \). Since \( \frac{2}{3} < 1 \), the series converges.

infinite series

An infinite series is the sum of an infinite sequence of terms. These can converge to a finite value or diverge to infinity. Understanding whether an infinite series converges is crucial in many areas of mathematics and its applications.
The general form of an infinite series is \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the sequence terms. In our exercise:

• The terms are given by \( \frac{n(n+1) \cdot 2^n}{3^n} \)

This series presents terms that grow and shrink through different components such as \( n(n+1) \) and \( \frac{2^n}{3^n} \). The overall behavior is eventually what determines convergence.

limit evaluation

Limit evaluation is a critical technique used alongside the Ratio Test to determine convergence. It involves calculating the behavior of a function or sequence as the input approaches a certain value, often infinity.
In the context of our series, we needed to evaluate:\[\lim_{n \to \infty} \frac{2(n+2)}{3n}\]This involves breaking down the expression so that direct substitution reflects the long-term behavior:

• Simplify the terms in the fraction to compare the dominant terms.
• The numerator \(2n + 4\) simplifies by recognizing that \(4\) becomes negligible as \(n\) grows very large.
• The result \(\frac{2n}{3n}\) simplifies to \(\frac{2}{3}\).

Reaching \( \frac{2}{3} \) tells us a lot about the convergence of the series.

factorial simplification

Factorial simplification is often a necessary step when dealing with terms in a series that include factorials, especially when using the Ratio Test.
In the given series problem, the factorial expression \( \frac{(n+1)!}{n!} \) allows direct simplification. This occurs by understanding:

• \((n+1)! = (n+1) \cdot n!\)
• So, \( \frac{(n+1)!}{n!} = n+1 \)

This step simplifies the series term significantly, making the entire process of applying the Ratio Test more manageable. Simplifying factorials often reduces the complexity of expressions so they can be compared or computed easily in further computations.