Problem 9 - Calculus: Early Transcendentals

Question

Use the Ratio Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 1}^{\infty} ( - 1)^{n-1} \frac {3^n}{2^n n^3} \)

Step-by-Step Solution

Answer

The series is divergent by the Ratio Test.

Step 1: Identify general term

The general term of the series can be written as \( a_n = (-1)^{n-1} \frac{3^n}{2^n n^3} \). For the Ratio Test, we can ignore the \((-1)^{n-1}\) factor since it does not affect the absolute value test.

Step 2: Setup Ratio Test

The Ratio Test involves finding the limit of the absolute value of the ratio of consecutive terms: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This simplifies to \( \lim_{n \to \infty} \frac{3^{n+1} / (2^{n+1} (n+1)^3)}{3^n / (2^n n^3)} \).

Step 3: Simplify the Ratio

Simplify the expression: \( \frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{2^{n+1}(n+1)^3} \times \frac{2^n n^3}{3^n} \). This further simplifies to \( \left(\frac{3}{2}\right) \cdot \frac{n^3}{(n+1)^3} \).

Step 4: Evaluate the Limit

Evaluate the limit: \[ \lim_{n \to \infty} \left(\frac{3}{2}\right) \cdot \frac{n^3}{(n+1)^3} \]. This simplifies to \( \frac{3}{2} \cdot \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^3 = \frac{3}{2} \cdot \lim_{n \to \infty} \left(\frac{n}{n}(1 + \frac{1}{n})\right)^3 = \frac{3}{2} \cdot 1 = \frac{3}{2} \).

Step 5: Conclusion from the Ratio Test

Since the limit \( \frac{3}{2} > 1 \), by the Ratio Test, the series \( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{3^n}{2^n n^3} \) is divergent.

Key Concepts

Convergence of a Series

Understanding whether a series converges or diverges is a critical part of mathematical analysis. A series converges if the sum of its infinite terms approaches a finite number as the number of terms increases sequentially. Conversely, a series diverges if its sum does not approach any single finite limit but instead grows without bound or oscillates without settling.
The series we are examining is \[ \sum_{n = 1}^{\infty} (-1)^{n-1} \frac{3^n}{2^n n^3}\]. The notation \(\sum_{n=1}^{\infty}\) indicates the sum starts at \(n=1\) and extends to infinity. The convergence or divergence of this series means we are investigating whether this infinite addition results in a specific number or grows infinitely/not definitively.
Using various tests such as the Ratio Test helps us determine convergence. When applied correctly, it indicates whether the absolute sum converges, which is often a precursor to determining convergence of the original series itself.

Limit Evaluation

The Ratio Test involves evaluating a limit to determine the convergence of a series. Computing limits is about finding what value a function approaches as the input approaches some point—in many cases, infinity.
In this case, to apply the Ratio Test, the goal was to evaluate the limit \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]. To find this limit, you simplify the expression involving the ratio of consecutive terms of a series. The expression is simplified by canceling out the common terms and rearranging the factors to a simpler form. The expression simplifies to \[ \left(\frac{3}{2}\right) \cdot \frac{n^3}{(n+1)^3} \].
As \(n\) approaches infinity, certain terms become negligible. Through simplifying \[ \frac{n}{n+1} \] become 1, the overall limit turns into \(\frac{3}{2} \cdot 1 = \frac{3}{2}\). Correctly evaluating these limits correctly dictates whether a series is convergent or divergent through the Ratio Test guidelines.

Absolute Value of a Series

When using the Ratio Test, the absolute value of a series plays a fundamental role. This is crucial when dealing with series that have terms capable of changing signs, like alternating series.
For this exercise, the initial general term is \[a_n = (-1)^{n-1} \frac{3^n}{2^n n^3}\]. The factor \((-1)^{n-1}\) causes alternating signs. However, for the Ratio Test, we consider the absolute value of the ratio of consecutive terms which means ignoring the negative signs.

• Absolute value ensures that we are only focusing on the magnitude of each term.
• In calculating the limit, this helps streamline what could otherwise be a more complicated problem.

Concentration on the magnitude via absolute values ensures the Ratio Test's applicability and provides clarity on the polynomial growth or decay, thus clearing the path to a straightforward conclusion about convergence.