Problem 69 - Single Variable Calculus

Question

Complete the following table to show that Equation (4). $$ e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} $$ appears to be valid. $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline n & 1 & 10 & 10^{2} & 10^{3} & 10^{4} & 10^{5} & 10^{6} \\ \hline\left(1+\frac{1}{n}\right)^{n} & & & & & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The completed table shows the values of \(\left(1+\frac{1}{n}\right)^{n}\) for different n: $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline n & 1 & 10 & 10^{2} & 10^{3} & 10^{4} & 10^{5} & 10^{6} \\ \hline\left(1+\frac{1}{n}\right)^{n} & 2 & 2.5937 & 2.7048 & 2.7169 & 2.7181 & 2.71827 & 2.71828 \\ \hline \end{array} $$ As n increases, the value of \(\left(1+\frac{1}{n}\right)^{n}\) converges towards e (approximately 2.71828), supporting the validity of the equation \(e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}\).

Step 1: Plug in n = 1 into the expression to find the value for the first column: \[ \left(1+\frac{1}{1}\right)^{1} = (1+1)^{1} = 2 \] #Step 2: Evaluate the expression for n=10#

Plug in n = 10 into the expression to find the value for the second column: \[ \left(1+\frac{1}{10}\right)^{10} \approx 2.5937 \] #Step 3: Evaluate the expression for n=\(10^2\)#

Step 2: Plug in n = \(10^2\) (100) into the expression to find the value for the third column: \[ \left(1+\frac{1}{100}\right)^{100} \approx 2.7048 \] #Step 4: Evaluate the expression for n=\(10^3\)#

Plug in n = \(10^3\) (1000) into the expression to find the value for the fourth column: \[ \left(1+\frac{1}{1000}\right)^{1000} \approx 2.7169 \] #Step 5: Evaluate the expression for n=\(10^4\)#

Step 3: Plug in n = \(10^4\) (10000) into the expression to find the value for the fifth column: \[ \left(1+\frac{1}{10000}\right)^{10000} \approx 2.7181 \] #Step 6: Evaluate the expression for n=\(10^5\)#

Plug in n = \(10^5\) (100000) into the expression to find the value for the sixth column: \[ \left(1+\frac{1}{100000}\right)^{100000} \approx 2.71827 \] #Step 7: Evaluate the expression for n=\(10^6\)#

Step 4: Plug in n = \(10^6\) (1000000) into the expression to find the value for the seventh column: \[ \left(1+\frac{1}{1000000}\right)^{1000000} \approx 2.71828 \] #Step 8: Complete the table with the calculated values#

Now that we have calculated the values of \(\left(1+\frac{1}{n}\right)^{n}\) for each value of n listed in the table, we can fill in the table: $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline n & 1 & 10 & 10^{2} & 10^{3} & 10^{4} & 10^{5} & 10^{6} \\ \hline\left(1+\frac{1}{n}\right)^{n} & 2 & 2.5937 & 2.7048 & 2.7169 & 2.7181 & 2.71827 & 2.71828 \\ \hline \end{array} $$ The table demonstrates that as n increases, the value of \(\left(1+\frac{1}{n}\right)^{n}\) appears to converge to e (approximately 2.71828), supporting the validity of the given equation: $$ e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} $$

Key Concepts

Convergence of Sequences

Understanding the convergence of sequences is a fundamental concept in calculus, as it relates to the behavior of sequences as their index approaches infinity. A sequence converges when its terms draw closer to a specific value as the index grows without bound. This is precisely what we observe with the sequence \( \left(1+\frac{1}{n}\right)^n \) as \( n \) approaches infinity—it approaches a specific number e, known as Euler's number. The exercise provided illustrates this by incrementally increasing the value of \( n \) and calculating the sequence's terms, which demonstrate steadily less deviation from \( e \).

Exponential Functions

The concept of exponential functions revolves around an equation where a constant base is raised to a variable exponent. These functions are crucial for modeling growth and decay processes in science and finance. The expression \( \left(1+\frac{1}{n}\right)^n \) from the exercise resembles the form of an exponential function, where the base \( \left(1+\frac{1}{n}\right) \) gets closer to 1 as \( n \) increases, but the exponent grows without limit. Despite its unconventional appearance, this sequence is a vital example of an exponential function's behavior at infinity and gives rise to the number \( e \), which is the base of natural logarithms and a cornerstone of continuous growth.

Limits in Calculus

The concept of limits in calculus is essential for analyzing functions' behaviors near specific points or at infinity. Limits help in understanding instantaneous rates of change (derivatives) and accumulated quantities (integrals). The expression \( e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \) encapsulates the limit definition of e and shows how it can be determined through the infinite process of taking higher and higher powers of sequences close to 1. This limit is not straightforward to evaluate by substitution since infinity is not a number, but a concept. Calculus tools such as L'Hôpital's rule or series expansion are often required for a rigorous proof, whereas the textbook exercise provides an intuitive numerical approach.

Infinite Series

While not directly illustrated by the exercise, the topic of infinite series is inherently linked to the limit definition of e. An infinite series is the sum of the terms of an infinite sequence. For example, the number \( e \) can also be represented as the sum of an infinite series: \( e = \sum_{n=0}^{\infty} \frac{1}{n!} \). This series, like the sequence in the table, converges to the numerical value of e as more terms are added. Understanding this and other series is significant for not only computing limits but also for developing deeper insights into functions and their behaviors in calculus.