Problem 100 - Calculus Early Transcendentals

Question

Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps. a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use \(\ln (a b)=\ln a+\ln b\) to show that $$ \begin{aligned} L &=\lim _{n \rightarrow \infty}\left(\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right) \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n} \end{aligned} $$ b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln x d x\) Integrate this improper integral by parts and reach the desired conclusion.

Step-by-Step Solution

Answer

Answer: The value of the limit is -1.

Step 1: Rewrite the limit using logarithm properties

Given \(L = \lim_{n \rightarrow \infty} \left(\frac{1}{n} \ln n! - \ln n\right)\). Using logarithm property, \(\ln(a b) = \ln a + \ln b\), and breaking down the logarithm of the factorial, we have: $$L = \lim_{n \rightarrow \infty}\left(\left(\frac{1}{n}\sum_{k=1}^{n} \ln k\right) - \ln n\right)$$

Step 2: Rewrite the limit as a Riemann sum

Using some algebra, we can rewrite the sum as a Riemann sum: $$L =\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n}$$

Step 3: Identify the Riemann sum as an improper integral

As \(n \rightarrow \infty\), the sum becomes the improper integral: $$L = \int_{0}^{1} \ln x dx$$

Step 4: Solve the integral using integration by parts

Now we will use integration by parts with \(u=\ln x\), \(dv=dx\) and thus, calculating derivatives and antiderivatives, we get \(du=\frac{1}{x}dx\) and \(v=x\). Applying integration by parts formula \(\int udv = uv - \int vdu\), we have: $$L = \left[x\ln x\right]_0^1 - \int_{0}^{1} x \frac{1}{x} dx = [-x]_0^1 = -1$$ Therefore, \(L = \lim_{n \rightarrow \infty}\left(\frac{1}{n} \ln n! - \ln n\right)=-1\).

Key Concepts

Improper Integral

When we deal with integrals that have infinite intervals or discontinuities, we refer to them as improper integrals. In our problem, the integral \[ L = \int_{0}^{1} \ln x \, dx \]is an improper integral due to the singularity at the lower limit, where \( x \) approaches zero. This integral requires careful handling because the natural logarithm \( \ln x \) tends to negative infinity as \( x \) approaches zero.

To effectively manage an improper integral like this, mathematicians apply limits to define the integral. For instance, instead of directly integrating from 0 to 1, we set up a limit as \( \epsilon \) approaches zero from the positive side (\( \epsilon^+ \)) so that: \[ \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1} \ln x \, dx. \]This essentially avoids directly dealing with the point of discontinuity, making it possible to compute the integral's value.

Factorial

The factorial function, denoted by an exclamation point \( n! \), is a key concept in permutations, combinations, and many areas of mathematics. It represents the product of an integer \( n \) multiplying all the integers down to 1. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow very quickly with larger \( n \), making them fundamental in functions such as the exponential and series expansions. In our exercise, we use logarithms to simplify the problem involving factorials into more manageable terms.

• Instead of dealing with large products directly, logarithms convert these products into sums.
• This step is crucial for rewriting the limit involving \( n! \) as a Riemann sum, which then relates back to the integral we need to evaluate.

Logarithmic Properties

Logarithmic properties are essential tools that allow us to simplify and transform complex mathematical expressions. Several key properties are important for our exercise:

• Product Rule: \( \ln(ab) = \ln a + \ln b \), which tells us that the logarithm of a product is the sum of the logarithms.
• Quotient Rule: \( \ln(a/b) = \ln a - \ln b \), which is useful when breaking apart fractions into simpler terms.
• Power Rule: \( \ln(a^b) = b \ln a \), simplifying logarithms of powers.

In our solution, the product rule is utilized to express the logarithm of a factorial \( n! \) as the sum \( \sum_{k=1}^{n} \ln k \). This transformation is pivotal in converting the problem into an equivalent form suitable for integration.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. It's useful in integrating products of functions, especially when standard methods might not work as efficiently. The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du. \]In our problem, we apply integration by parts to solve the improper integral \( \int_{0}^{1} \ln x \, dx \).

• Choose \( u = \ln x \) and \( dv = dx \). This choice is essential because differentiating \( \ln x \) simplifies the problem while integrating \( dx \) remains straightforward.
• With these choices, differentiate to get \( du = \frac{1}{x} dx \) and integrate \( dv \) to get \( v = x \).

Substitute into the integration by parts formula: \[ \int \ln x \, dx = \left[x \ln x \right]_0^1 - \int_{0}^{1} x \frac{1}{x} \, dx. \]Solve the remaining integral to arrive at the final answer, handling the limits carefully, especially as \( x \to 0^+ \), yielding the solution \( L = -1 \). This powerful technique transforms complex integrals into simpler components, making it a staple in calculus.