Problem 8 - Thomas Calculus

Question

In Exercises \(1-8,\) use the Ratio Test to determine if each series converges absolutely or diverges. $$ \sum_{n=1}^{\infty} \frac{n 5^{n}}{(2 n+3) \ln (n+1)} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Understand the Ratio Test

The Ratio Test states that for a series \( \sum a_n \), if the limit \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]exists and is equal to a value \(L\):1. If \( L < 1 \), then the series converges absolutely.2. If \( L > 1 \), then the series diverges.3. If \( L = 1 \), the test is inconclusive.

Step 2: Identify Terms for the Ratio Test

For the given series \( \sum_{n=1}^{\infty} \frac{n 5^{n}}{(2 n+3) \ln (n+1)} \), the general term can be identified as:\[ a_n = \frac{n 5^{n}}{(2 n + 3) \ln (n + 1)} \] Next, find the expression for \( a_{n+1} \):\[ a_{n+1} = \frac{(n+1) 5^{n+1}}{(2(n+1) + 3) \ln((n+1) + 1)} \] This simplifies to \[ a_{n+1} = \frac{(n+1) 5^{n+1}}{(2n + 5) \ln(n + 2)} \].

Step 3: Compute the Ratio \( \left| \frac{a_{n+1}}{a_n} \right| \)

Substitute \( a_{n+1} \) and \( a_n \) into the ratio:\[ \frac{a_{n+1}}{a_n} = \frac{(n+1) 5^{n+1}}{(2n + 5) \ln(n + 2)} \cdot \frac{(2n + 3) \ln(n + 1)}{n 5^n} \] Simplify:\[ \frac{a_{n+1}}{a_n} = \frac{(n+1) 5 \cdot (2n + 3) \ln(n + 1)}{n (2n + 5) \ln(n + 2)} \].

Step 4: Simplify and Find the Limit \( L \)

Further simplify the expression:\[ \frac{a_{n+1}}{a_n} = \frac{5 (n+1) (2n+3) \ln(n+1)}{n (2n+5) \ln(n+2)} \] This expression needs to be simplified as \( n \to \infty \):- Note that the dominant factors for large \( n \) are \( n \), so the expression simplifies to:\[ \lim_{n \to \infty} \frac{5(n+1)}{n} \cdot \frac{2n+3}{2n+5} \cdot \frac{\ln(n+1)}{\ln(n+2)} \]- As \( n \to \infty \), the limits of each factor become: - \( \frac{n+1}{n} \to 1 \) - \( \frac{2n+3}{2n+5} \to 1 \) - \( \frac{\ln(n+1)}{\ln(n+2)} \to 1 \)Thus, the entire limit \( L = 5 \times 1 \times 1 \times 1 = 5 \).

Step 5: Determine Convergence or Divergence

From the Ratio Test, since \( L = 5 > 1 \), the series \( \sum_{n=1}^{\infty} \frac{n 5^{n}}{(2 n+3) \ln (n+1)} \) diverges.

Key Concepts

Series Convergence

A series is said to converge if the sum of its terms approaches a specific finite value as more terms are added. Imagine starting with a number and continuously adding smaller and smaller numbers to it, leading to a stable sum. That's essentially what convergence means. In more mathematical terms, for a series \( \sum_{n=1}^{\infty} a_n \), convergence means the limit \( \lim_{N \to \infty} \sum_{n=1}^{N} a_n \) exists and is finite.
However, identifying convergence can be challenging. That is where tests like the Ratio Test become immensely helpful. They provide a systematic way to assess whether a series converges. Using the Ratio Test, if \( L < 1 \), the series converges absolutely, meaning not only does the sum exist, but it can withstand some manipulation like changing the order of its terms without affecting the outcome.
If you apply this knowledge while solving problems like the example series \( \sum_{n=1}^{\infty} \frac{n 5^{n}}{(2 n+3) \ln (n+1)} \), correctly identifying \( L \) can straightforwardly tell you about the series' behavior.

Divergence of Series

When a series diverges, it does not approach any specific value, no matter how many terms are added. Instead, it might continue growing indefinitely or oscillate without settling down. For a series \( \sum_{n=1}^{\infty} a_n \), divergence is the result when the limit \( \lim_{N \to \infty} \sum_{n=1}^{N} a_n \) does not exist or is infinite.
In the realm of the Ratio Test, if the value \( L > 1 \), it clearly indicates that the series diverges. This is exactly what happens with our example exercise: after determining the limit \( L \), it was found that \( L = 5 \), which is greater than 1, confirming the divergence of the series \( \sum_{n=1}^{\infty} \frac{n 5^{n}}{(2 n+3) \ln (n+1)} \).
Intuitively, imagine building a tower by stacking progressively larger blocks. If the size of your blocks doesn't shrink fast enough, your tower (i.e., the series sum) will just keep rising higher without stabilizing, akin to divergence.

Limit of a Sequence

Understanding the limit of a sequence is crucial as it's the foundation for dealing with series. A sequence is simply an ordered list of numbers, like \( a_1, a_2, a_3, \ldots \). The limit of a sequence describes what value a sequence approaches as you move further and further along it. In mathematical terms, a sequence \( a_n \) converges to \( L \) if \( \lim_{n \to \infty} a_n = L \).
This concept is especially important within the context of the Ratio Test. When determining convergence using the Ratio Test, we compute the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If the limit \( L \) is less than 1, the series is assuredly convergent. If greater than 1, divergence is certain. In this practical application, understanding how to compute and interpret these limits unlocks insights into series behaviors.
Returning to our original exercise, computing the limit of the sequence formed by \( \left| \frac{a_{n+1}}{a_n} \right| \) gave us \( L = 5 \). Recognizing this ensured our understanding of the series as diverging, guided by this principle of sequence limits.