Problem 33 - Calculus: One and Several Variables

Question

Show that $$ \sum_{x=1}^{\infty} \ln \left(\frac{k+1}{k}\right) \quad \text { diverges } $$ although $$ \ln \left(\frac{k}{k}-\frac{1}{k}\right) \rightarrow 0 $$

Step-by-Step Solution

Answer

To show that the sum $$\sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right)$$ diverges, we use logarithm properties to rewrite the sum as the natural logarithm of a product: $$\ln \left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \right) = \ln(n!)$$ Since the factorial function grows much faster than exponential functions, its natural logarithm also grows without bound. Therefore, the sum diverges. This result is consistent with the given limit $$\ln \left(\frac{k}{k}-\frac{1}{k}\right) \rightarrow 0$$, showing that the behavior of the terms in a series and the behavior of their sum can be very different.

Step 1: Write out the sum of the series

We are studying the sum of each of the terms, written out as a series: $$ \sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right) $$#

Step 2: Remember logarithm properties

Recall that the sum of logarithms is equal to the logarithm of the product. Applying this property to our series, we get: $$ \ln \left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \right) $$

Step 3: Simplify the product

Notice that the terms in the product cancel out, except for the first term in the numerator and the last term in the denominator: $$ \ln \left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \right) = \ln(1 \cdot 2 \cdot 3 \cdots n) $$

Step 4: Analyze the product

We see that the product is the factorial of n, denoted as n! This can be written as: $$ \ln(n!) $$ Since the factorial function grows much faster than exponential functions, we know that its natural logarithm also has to grow without bound.

Step 5: State the result

Given that the natural logarithm of the factorial grow withoud bound, we can conclude that: $$ \sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right) \quad \text { diverges } $$

Step 6: Discuss the limit

It is also given that: $$ \ln \left(\frac{k}{k}-\frac{1}{k}\right) \rightarrow 0 $$ as k goes to infinity. This is consistent with our result since the sum of logarithms diverges even though the individual terms approach 0. This shows that the behavior of the terms in a series and the behavior of their sum can be very different.

Key Concepts

Logarithmic Properties

The properties of logarithms are powerful tools when evaluating series or solving mathematical problems related to growth rates and limits. One important property is that the logarithm of a product is the sum of the logarithms of the factors. For example:

• \( \ln(a \cdot b) = \ln a + \ln b \)

This property allows us to break down complex expressions into much simpler forms. In the context of series, applying this principle can help identify patterns or cancellations within the terms.

In the exercise, the series \( \sum_{k=1}^{\infty} \ln \left(\frac{k+1}{k}\right) \) simplifies due to telescoping effect, leading to a product that concerns consecutive numbers in a sequence. Understanding that this is possible by the logarithmic product rule is crucial in finding that many terms cancel out, yielding a remarkably concise form. Thus, using logarithmic properties effectively can simplify the understanding of convergence or divergence within series.

Factorial Growth

Factorials are a fundamental concept in mathematics, representing the product of all positive integers up to a given number \( n \). The notation \( n! \) formalizes this definition. The factorial function grows exceptionally fast as \( n \) increases.

Even faster than exponential functions, this rapid escalation becomes particularly evident in large scale calculations, such as permutations and combinations in probability, or when simplifying the terms of a series. For example:

• \( 5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 \)

Factorial growth plays a critical role in evaluating the convergence of series because the sequence grows too rapidly to result in a finite sum. When considered with logarithms, as seen in the exercise, \( \ln(n!) \) grows without bounds, leading the series to diverge. Understanding factorial growth helps us realize why certain series, even those with decreasing terms, do not converge.

Limit Behavior

The behavior of limits is essential in advanced mathematics, providing insights into the long-term behavior of sequences or functions. A limit describes what a function approaches as the input approaches a certain value. Specifically, if a sequence \( a_k \to 0 \) as \( k \to \infty \), it does not necessarily imply the associated series \( \sum a_k \) converges.

In our exercise, analyzing individual term behavior \( \ln \left(\frac{k}{k}-\frac{1}{k}\right) \rightarrow 0 \) as \( k \to \infty \) exhibits how individual terms can approach zero without leading to convergence of the series. Such phenomena illustrate that a sequence tending towards zero may still lead to a divergent series, typically due to insufficiently rapid decay of terms. This underscores the distinction between the limit of terms and the sum of the series, emphasizing that convergence criteria for series are more stringent than for sequences.