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 $$ \begin{aligned} L &=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right] \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right) \end{aligned} $$ 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.

Short answer

Question: Show that the following limit is true: \(\lim_{n\rightarrow\infty} \left(\frac{1}{n}\ln n! - \ln n\right) = -1\). Solution: By rewriting the given expression as a Riemann sum, identifying the corresponding improper integral, evaluating it using integration by parts, and taking the limit of the integral, we have shown that the given limit equals -1.

Step-by-step solution

Rewrite the expression as a Riemann sum

: Given \(L = \lim_{n\rightarrow\infty} \left(\frac{1}{n}\ln n! - \ln n\right)\), let's start by observing that \(n! = n(n-1)(n-2)\cdots1\) and use the logarithmic property that \(\ln(ab) = \ln a + \ln b\). $$ \begin{aligned} L &= \lim_{n\rightarrow\infty} \left(\frac{1}{n}\ln n! - \ln n\right) \\ &= \lim_{n\rightarrow\infty} \left(\frac{1}{n}\sum_{k=1}^{n}\ln k - \ln n\right) \\ &= \lim_{n\rightarrow\infty} \left[\left(\frac{1}{n}\sum_{k=1}^{n}\ln k\right) - \ln n\right]\\ &= \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{k}{n}\right) \end{aligned} $$

Identify the Riemann sum

: Now, let's observe that the given expression represents a Riemann sum for the improper integral \(\int_{0}^{1}\ln x\ dx\). To see this, recall that a Riemann sum is given by \(\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x\), where \(x_i^*\) are points in the subintervals and \(\Delta x = \frac{b-a}{n}\). In this case, \(a=0\), \(b=1\), \(\Delta x = \frac{1}{n}\), and \(x_i^* = \frac{k}{n}\), hence $$ L = \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{k}{n}\right) = \int_{0}^{1} \ln x\ dx $$

Integrate by parts

: To evaluate the improper integral, we will use integration by parts. Recall that integration by parts is given by \(\int u\ dv = uv - \int v\ du\). Let \(u=\ln x\) and \(dv=dx\). Differentiate \(u\) and integrate \(dv\) to get \(du=\frac{1}{x}dx\) and \(v=x\). Now apply the integration by parts formula: $$ \begin{aligned} \int_{0}^{1} \ln x\ dx &= \int u\ dv \\ &= uv - \int v\ du \\ &= x\ln x - \int x\left(\frac{1}{x}\right) dx \\ &= x\ln x - \int dx \\ &= x\ln x - x \end{aligned} $$

Evaluate the limit of the integral

: Finally, let's evaluate the improper integral by taking the limit as the bounds approach 0 and 1: $$ \begin{aligned} L &= \lim_{n\rightarrow\infty}\left(\frac{1}{n}\sum_{k=1}^{n}\ln\frac{k}{n}\right) \\ &= \int_{0}^{1} \ln x\ dx \\ &= \left[\lim_{x\rightarrow 0^+}(x \ln x - x)\right]-\left[\lim_{x\rightarrow 1^-}(x \ln x - x)\right] \\ &= (0 - 0) - (1\cdot 0 - 1) \\ &= -1 \end{aligned} $$ Therefore, we have shown that \(L=\lim_{n\rightarrow\infty}\left(\frac{1}{n}\ln n!-\ln n\right)=-1\).

Key concepts

Riemann Sum

Riemann sums are a key concept in calculus used to approximate the total area underneath a curve. It's a method that breaks the area into small rectangles whose areas are easy to add up. The sum of these rectangles approaches the exact area as the width of the rectangles decreases.
To understand the Riemann sum in our problem, consider the expression \( \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right) \). This is a classic example of how Riemann sums approximate an integral.

• Divide the interval from 0 to 1 into \( n \) equal parts.
• The function we are integrating is \( \ln x \), evaluated at sample points \( \frac{k}{n} \).
• As \( n \) increases, the approximation becomes more accurate, mathematically approaching the integral \( \int_{0}^{1} \ln x\, dx \).

This approach vividly illustrates how Riemann sums can transform complex problems into more manageable forms by using simple geometric concepts.

Improper Integral

Improper integrals are used when integrating over an interval that is unbounded or where the function has an infinite discontinuity. For our case, \( \int_{0}^{1} \ln x\, dx \) needs careful evaluation because \( \ln x \) tends towards minus infinity as \( x \to 0^+ \).

• This is why it is called ‘improper’—the limit has to be taken to evaluate it properly.
• To compute this integral, it’s essential first to transform it into a limit problem.

The core idea is to evaluate the integral from some positive \( \varepsilon \) to 1, and then take the limit as \( \varepsilon \to 0^+ \). This ensures that we are carefully managing the behavior of \( \ln x \) near zero so the solution can progressively encompass the entire range from 0 to 1.

Integration by Parts

Integration by parts is a powerful technique used to integrate products of functions. It is based on the product rule for differentiation and reverses the process of differentiation under the integral sign.

• The formula \( \int u\, dv = uv - \int v\, du \) is used here effectively.

In the context of our problem, we set \( u = \ln x \) and \( dv = dx \). Differentiating and integrating these terms gives us \( du = \frac{1}{x} dx \) and \( v = x \).

• Plug these values into the integration by parts formula to rework the integral.
• Compute the residual integral \( \int v\, du = \int 1\, dx \) which simplifies to \( x \).

Through these steps, we simplify and evaluate the original integral \( \int_{0}^{1} \ln x\, dx \). This method effectively leverages the interaction between the two chosen functions, allowing us to evaluate otherwise challenging integrals.