Problem 8 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series diverges by the Ratio Test.

Step 1: Identify the Series Terms

The given series is \( \sum_{n=1}^{\infty} \frac{n \cdot 5^{n}}{(2n+3) \ln(n+1)} \). Identify the general term of the series, which is \( a_n = \frac{n \cdot 5^{n}}{(2n+3) \ln(n+1)} \).

Step 2: Apply the Ratio Test Formula

The Ratio Test considers the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Substitute the general terms for \( a_{n+1} \) and \( a_n \).

Step 3: Compute \( a_{n+1} \)

Compute \( a_{n+1} = \frac{(n+1)5^{n+1}}{(2(n+1)+3) \ln((n+1)+1)} = \frac{(n+1)5^{n+1}}{(2n+5) \ln(n+2)} \).

Step 4: Formulate the Ratio \( \frac{a_{n+1}}{a_n} \)

Calculate the ratio:\[\frac{a_{n+1}}{a_n} = \frac{(n+1)5^{n+1}}{(2n+5)\ln(n+2)} \times \frac{(2n+3)\ln(n+1)}{n \cdot 5^n}\]Simplify to:\[\frac{a_{n+1}}{a_n} = 5 \cdot \frac{n+1}{n} \cdot \frac{2n+3}{2n+5} \cdot \frac{\ln(n+1)}{\ln(n+2)}\]

Step 5: Calculate the Limit of the Ratio

Take the limit as \( n \to \infty \):\[L = \lim_{n \to \infty} \left( 5 \cdot \frac{n+1}{n} \cdot \frac{2n+3}{2n+5} \cdot \frac{\ln(n+1)}{\ln(n+2)} \right)\]Evaluate each part, noticing that \( \frac{n+1}{n} \to 1 \), \( \frac{2n+3}{2n+5} \to 1 \), and \( \frac{\ln(n+1)}{\ln(n+2)} \to 1 \), so:\[ L = 5 \times 1 \times 1 \times 1 = 5\]

Step 6: Interpret the Result of the Ratio Test

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

Key Concepts

Series Convergence

Series convergence is a crucial concept when studying infinite series. An infinite series is simply the sum of an infinite sequence of numbers. Whether a series converges means checking if, as you keep adding more and more terms, the sum approaches a specific finite number. If it does, the series is said to converge.
Different tests can be used to determine convergence, such as the Ratio Test, Root Test, or the Integral Test. In practical exercises, like the one we're looking at, the Ratio Test is often applied. If the test shows a result less than 1, the series converges absolutely. However, if it's greater than or equal to 1, the series diverges or converges conditionally.
The convergence of an infinite series is essential in various mathematical fields, including calculus, to assure solutions behave as expected under certain conditions.

Limit of a Sequence

The limit of a sequence is foundational when dealing with series convergence. Even though sequences themselves are not infinite sums, understanding their behavior at infinity helps judge series. A sequence of numbers approaches a limit if the elements of the sequence get closer to a specific number as the number of terms grows.
The Ratio Test uses limits to determine convergence by analyzing the ratio of successive terms in a series. This means calculating:

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)

If this ratio stabilizes to less than 1, the series is deemed convergent.
In the original problem, calculating the limit involves simplifying the expression for large \( n \), focusing on which terms grow the fastest to determine convergence behavior.

Divergence

Divergence suggests that as you sum more terms, the total does not settle to a fixed number, but instead grows indefinitely or oscillates without reaching any finite value. Understanding divergence is as vital as convergence because it tells us when an infinite series cannot be assigned a finite sum.
Using the Ratio Test involves a key aspect of divergence: when the calculated limit \( L \) is greater than 1, it confirms that the series diverges. This was the case in our original exercise, where the calculated \( L = 5 \) determined that the series could not converge.
Recognizing divergence effectively helps in identifying when a problem or solution might require a different approach, such as looking for other ways to manage or transform the problem space.