Problem 5 - Active Calculus

Question

Finding limits of convergent sequences can be a challenge. However, there is a useful tool we can adapt from our study of limits of continuous functions at infinity to use to find limits of sequences. We illustrate in this exercise with the example of the sequence $$ \frac{\ln (n)}{n} $$ a. Calculate the first 10 terms of this sequence. Based on these calculations, do you think the sequence converges or diverges? Why? b. For this sequence, there is a corresponding continuous function \(f\) defined by $$ f(x)=\frac{\ln (x)}{x} $$ Draw the graph of \(f(x)\) on the interval [0,10] and then plot the entries of the sequence on the graph. What conclusion do you think we can draw about the sequence \(\left\\{\frac{\ln (n)}{n}\right\\}\) if \(\lim _{x \rightarrow \infty} f(x)=L ?\) Explain. c. Note that \(f(x)\) has the indeterminate form \(\frac{\infty}{\infty}\) as \(x\) goes to infinity. What idea from differential calculus can we use to calculate \(\lim _{x \rightarrow \infty} f(x) ?\) Use this method to find \(\lim _{x \rightarrow \infty} f(x) .\) What, then, is \(\lim _{n \rightarrow \infty} \frac{\ln (n)}{n} ?\)

Step-by-Step Solution

Answer

The sequence \( \frac{\ln(n)}{n} \) converges to 0 as \( n \) approaches infinity.

Step 1: Calculate the first 10 terms

To calculate the first 10 terms of the sequence \ \(\frac{\ln(n)}{n}\), substitute \(n\) with values from 1 to 10. The natural logarithm \(\ln(n)\) of each value of \(n\) can be found using a calculator or logarithm table. Then divide by \(n\) to get the terms. \[ \begin{align*} &n = 1, \ \frac{\ln(1)}{1} = 0, \ &n = 2, \ \frac{\ln(2)}{2} \approx 0.3466, \ &n = 3, \ \frac{\ln(3)}{3} \approx 0.3662, \ &n = 4, \ \frac{\ln(4)}{4} \approx 0.3466, \ &n = 5, \ \frac{\ln(5)}{5} \approx 0.3219, \ &n = 6, \ \frac{\ln(6)}{6} \approx 0.2986, \ &n = 7, \ \frac{\ln(7)}{7} \approx 0.2773, \ &n = 8, \ \frac{\ln(8)}{8} \approx 0.2591, \ &n = 9, \ \frac{\ln(9)}{9} \approx 0.2427, \ &n = 10, \ \frac{\ln(10)}{10} \approx 0.2303. \end{align*} \] Based on these calculations, we notice that the sequence values are getting smaller. It appears that the terms of the sequence are approaching 0 as \(n\) increases. Thus, the sequence seems to converge.

Step 2: Define the continuous function and plot the graph

The continuous function \(f(x)\) corresponding to the sequence is given by \[f(x) = \frac{\ln(x)}{x}\] Plot the graph of \(f(x)\) on the interval \([0, 10]\). Notice that \(f(x)\) is undefined at \( x=0 \) since \( \ln(0) \) is undefined. Then, plot the sequence points \( \left\{\frac{\ln(n)}{n}\right\} \) for \( n=1, 2, 3, \, ... , \, 10 \). The graph reveals that as \( x \) increases, \( f(x) \) seems to approach 0. This suggests that \( \lim_{x \to \infty} f(x) = 0 \). Consequently, \(\lim_{n \to \infty} \frac{\ln(n)}{n} = 0\).

Step 3: Use L'Hôpital's Rule to find the limit

Since the formula \( \frac{\ln(x)}{x} \) is an indeterminate form \( \frac{\infty}{\infty} \) as \( x \to \infty \), apply L'Hôpital's Rule to find the limit. L'Hôpital's Rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is indeterminate, then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) provided the latter limit exists. \[\begin{align*} &f(x) = \ln(x), \, f'(x) = \frac{1}{x} \ &g(x) = x, \, g'(x) = 1 \ &\text{Therefore,} \lim_{x \to \infty} \frac{\ln(x)}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx} \ln(x)}{\frac{d}{dx} x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0 \end{align*} \] Hence, \( \lim_{n \to \infty} \frac{\ln(n)}{n} = 0 \).

Key Concepts

Convergence

In mathematics, convergence refers to the idea that a sequence or series approaches a specific value as its terms progress towards infinity. For a sequence \(a_n\) to converge, there exists a limit \(L\) such that for any smaller-than-desired distance, all subsequent terms of the sequence are within that distance from \(L\). For instance, in our exercise, after calculating the first 10 terms of the sequence \( \frac{\ln(n)}{n} \), we observe a pattern: \0, 0.3466, 0.3662, 0.3466, 0.3219, 0.2986, 0.2773, 0.2591, 0.2427, 0.2303\. We see the terms are decreasing.

This observation gives a hint that they might be approaching 0 as \(n\) becomes very large. Mathematical proof later confirmed this hypothesis, showing that \( \lim_{n \to \infty} \frac{\ln(n)}{n} = 0\). Thus, understanding how to recognize and prove convergence is key to mastering sequences.

Difference between sequences and functions

A sequence is an ordered list of numbers, typically defined by a formula for its general term. Each term is indexed by an integer and the sequence is discrete. On the other hand, a function is a relation that assigns to each input exactly one output, and can be continuous.

The primary difference lies in their domains: sequences are defined on the set of natural numbers, while functions can be defined on real numbers, allowing for values in between integers.

In the exercise, we transformed the sequence \( \frac{\ln(n)}{n} \) into a function \( f(x) = \frac{\ln(x)}{x} \) and analyzed its behavior. This allowed us to use tools from calculus, like L'Hôpital's Rule, for finding limits that might be more difficult to evaluate when treated as a sequence.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in differential calculus for evaluating limits involving indeterminate forms. If a limit is in the form \(\frac{\infty}{\infty} \) or \(\frac{0}{0} \), L'Hôpital's Rule states that \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right-hand side exists.

In the provided exercise, the function \(\frac{\ln(x)}{x} \) as \(x\) approaches infinity forms an indeterminate \(\frac{\infty}{\infty} \). We applied L'Hôpital's Rule by differentiating the numerator and the denominator:
\[ \lim_{x \to \infty} \frac{\ln(x)}{x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0 \]
This confirmed that our sequence \( \frac{\ln(n)}{n} \) converges to 0 as \( n \to \infty. \). Thus, mastering L'Hôpital's Rule can simplify finding limits in cases involving complex ratios.