Problem 100 - Essential Calculus

Question

(a) Show that \(\int_{1}^{n} \ln x d x<\ln (n !)\) for \(n \geq 2\). (b) Draw a graph similar to the one above that shows \(\ln (n !)<\int_{1}^{n+1} \ln x d x\) (c) Use the results of parts (a) and (b) to show that \(\frac{n^{n}}{e^{n-1}}1\) (d) Use the Squeeze Theorem for Sequences and the result of part (c) to show that \(\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n !}}{n}=\frac{1}{e}\) (e) Test the result of part (d) for \(n=20,50,\) and 100 .

Step-by-Step Solution

Answer

After applying the analyses and calculations, can confirm that the inequality \( \int_{1}^{n} \ln x d x < \ln (n !) \) for \( n \geq 2 \) holds true. Also, \( \ln (n !) < \int_{1}^{n+1} \ln x d x \) was shown graphically. The inequalities \( \frac{n^{n}}{e^{n-1}}1 \) was proven using these results. Finally, by using these inequalities and the Squeeze Theorem, it has been found that \( \lim _{n \rightarrow \infty} \frac{\sqrt[n]{n !}}{n}=\frac{1}{e} \). This result has been tested for \( n = 20, 50, 100 \) and the values were found to be approaching \( \frac{1}{e} \).

Step 1: Prove Integral Inequality

Observing the graphical representation of \( \int_{1}^{n} \ln x d x \) and \( \ln (n !) \), we can observe that the area under the curve of \( \ln x \) is always less than the sum of the areas of the rectangles whose heights are \( \ln 2, \ln 3, \ldots, \ln n \). Hence, we can conclude that \( \int_{1}^{n} \ln x d x < \ln (n !) \) for \( n>=2 \)

Step 2: Draw a Graph

The graph should show that the sum \( \ln (n !) \) is less than the integral from 1 to \( n+1 \) of \( \ln x dx \). The sum of the areas of the rectangles would be \( \ln (2) + \ln (3) + \cdots + \ln (n) \), which is \( \ln (n!) \). However, increasing the upper limit of integration to \( n+1 \) allows the integral area to exceed the sum, demonstrating \( \ln (n !) < \int_{1}^{n+1} \ln x dx \).

Step 3: Prove the exponential factorial inequalities

Known that \( \int_{1}^{n} \ln x d x < \ln (n !) < \int_{1}^{n+1} \ln x dx \), It's necessary to prove that \( \int_{1}^{n} \ln x d x = n \ln n - n +1 \) and \( \int_{1}^{n+1} \ln x dx = (n+1) \ln (n+1) - (n+1) +2 \). Using these equalities, we can rewrite the inequalities like: \( n \ln n - n +1 < \ln (n !) < (n+1) \ln (n+1) -(n+1) +2 \). Applying exponentiation results in the desired inequalities \( \frac{n^{n}}{e^{n-1}}1 \).

Step 4: Use the Squeeze Theorem

It can be rewritten the inequalities as \( \frac{n^n}{e^n} < n! < \frac{(n+1)^{n+1}}{e^n} \). Then take the nth root of each part gives \( \frac{n}{e}<\sqrt[n]{n !}<\frac{(n+1)(n+1)^{n}}{e^n} \). Dividing by n on each part we get \( \frac{1}{e}<\frac{\sqrt[n]{n !}}{n}<\frac{(n+1)^n}{e^n} \) which as \( n \rightarrow \infty \) both end go to \( \frac{1}{e} \) hence by squeeze theorem \( \lim _{n \rightarrow \infty} \frac{\sqrt[n]{n !}}{n}=\frac{1}{e} \)

Step 5: Verify the Limit

We verify this limit by plugging in the values \( n = 20, 50, 100 \). For each of these values, evaluate \( \frac{\sqrt[n]{n !}}{n} \). This value should approach \( \frac{1}{e} \) as \( n \) increases.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies continuous change, and it is divided into two major areas: differential calculus and integral calculus. Differential calculus concerns itself with the concept of the derivative, which shows how a function changes at any given point. Integral calculus, on the other hand, is about the integral, which can be perceived as the accumulation of quantities over an interval.

In the context of the given exercise, integral calculus is used to compare the growth of functions by examining the areas under curves. The inequality (t_{1}^{n} n x d x< n (n !)) stems from recognizing that the area under the logarithmic function, ( n x), from 1 to n, is less than the sum of the rectangles formed by consecutive values of this function, which are related to factorial growth. The logarithm and the integral are deeply intertwined in calculus because the logarithm function ( n x) has a natural connection with rates of growth, which can be explored through integration.

Factorial function

The factorial function, denoted by n!, is fundamental in both calculus and combinatorics. It represents the product of all positive integers up to a given number n. For example, 5! = 5 imes 4 imes 3 imes 2 imes 1 = 120. As numbers grow larger, their factorials increase extremely rapidly, surpassing exponential growth for sufficiently large values of n.

Factorials are related to permutations and combinations, reflecting the total counts of ways to order or select items. In calculus, the factorial appears in series expansions, such as the Taylor series, where variables raised to a power n are divided by n! providing critical approximations for complex functions. In the exercise, factorials play a key role in establishing bounds for n!, enabling a deep exploration into the behavior of sequences as n tends toward infinity, and demonstrating the growth rate of factorials in relation to exponentials.

Squeeze theorem

The Squeeze Theorem, also known as the Sandwich Theorem or the Pinching Theorem, is an important result in calculus that gives a method for finding the limit of a function. The principal idea is simple: if a function is 'squeezed' between two other functions that each approach the same limit at a specific point, then the 'squeezed' function must also approach the same limit at that point.

Specifically, if f(x) leq g(x) leq h(x) and the limits of f(x) and h(x) as x approaches a certain value are the same, then the limit of g(x) as x approaches that value must be the same as well. This theorem is particularly helpful when it's hard to evaluate the limit of a function directly, but it's easier to recognize its bounds. In the problem we are addressing, the Squeeze Theorem is cleverly used to show that the limit of the nth root of n factorial divided by n equals (1/e) as n grows without bounds, reinforcing the profound connections within calculus concepts.

Limits of sequences

Limits of sequences are a foundational concept in calculus and analysis, referring to the value that a sequence trends towards as the index (n) goes to infinity.

If a sequence (a_n) has a limit L, we can say that for any arbitrarily small number (epsilon), there exists a natural number N such that for all n > N, the distance between a_n and L is less than epsilon. Limits can often be tricky to calculate directly, but many theorems, such as the Squeeze Theorem, help in establishing these limits.

In our exercise, the limit of the sequence (ts[n]{n !}/n) as n approaches infinity is examined. It is a beautiful illustration of how sequences behave and how their limits provide profound insights into the nature of mathematical growth, particularly when factorial functions are involved. Through the Squeeze Theorem, we find that this sequence approaches (1/e), demonstrating a balance between the rapid growth of factorials and the moderation introduced by taking nth roots and dividing by n.