Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Show that $$ \lim _{n \rightarrow \infty} 2^{n}=\infty $$ using the following procedure. Argue using mathematical induction that all of the statements in the sequence \(\left\\{S_{1}, S_{2}, S_{3}, \cdots\right\\}\) of statements are true where \(S_{n}\) is the statement that \(n<2^{n},\) for \(n=1,2,3, \cdots\) Your argument should have two parts. Part \(1 .\) Show that \(S_{1}\) is true. Part 2 . If one of the statements in \(\left\\{S_{1}, S_{2}, S_{3}, \cdots\right\\}\) is not true, there must be a first one that is not true. Let \(m\) be the subscript of the first statement in \(\left\\{S_{1}, S_{2}, S_{3}, \cdots\right\\}\) that is not true. Show that 1\. \(1

Step-by-Step Solution

Answer

By induction, \(n < 2^n\) for all \(n\), so \(\lim_{n \to \infty} 2^n = \infty\) and \(\lim_{n \to \infty} 2^{-n} = 0\).

Step 1: Verify S1

Start by proving that the statement \(S_1\) is true. For \(n=1\), we need to check if \(1 < 2^1\). Clearly, \(1 < 2\), so \(S_1\) is true.

Step 2: Assume Inductive Hypothesis

Assume that \(S_k\) is true for some arbitrary positive integer \(k\). This means \(k < 2^k\).

Step 3: Prove for S_k+1

We need to show that if \(S_k\) is true, then \(S_{k+1}\) is also true. Consider \(k+1 < 2^{k+1}\). Since \(k < 2^k\) by the inductive hypothesis, multiplying by 2 gives \(2k < 2^{k+1}\). Since \(k+1 \leq 2k\) for \(k \geq 1\), it follows that \(k+1 < 2^{k+1}\). Thus, \(S_{k+1}\) holds.

Step 4: Conclude by Induction

By mathematical induction, since \(S_1\) is true and \(S_k\) implies \(S_{k+1}\), all statements \(S_n\) are true for all \(n\geq 1\).

Step 5: Implications of Induction

Since \(n < 2^n\) holds for all \(n\), it means \(2^n\) grows faster than any constant sequence, leading \(\lim_{n \to \infty} 2^n = \infty\). Since \(e^n\) grows even faster than \(2^n\), \(\lim_{n \to \infty} e^n = \infty\) as well.

Step 6: Conclude for 2^-n

Since \(2^n\) grows to infinity as \(n\) approaches infinity, \(2^{-n}\) tends towards zero. Therefore, \(\lim_{n \to \infty} 2^{-n} = 0\).

Key Concepts

Limits in Calculus

The concept of limits is fundamental in calculus, representing the value that a function or sequence 'approaches' as the input approaches some value. In this exercise, we are considering the limit of an exponential function as the term number, denoted by \( n \), approaches infinity. This is written as \( \lim_{n \to \infty} 2^n = \infty \), which implies that the sequence grows without bound.
When dealing with limits, especially involving infinities, it is crucial to understand that the function is unbounded; it can grow larger than any pre-set number. This implies that for any large number \( M \), there exists a point in the sequence where all subsequent numbers are larger than \( M \). In our example, this means that for any chosen value, however large, the function \( 2^n \) will eventually exceed this value as \( n \) increases.
This understanding of limits is essential for evaluating behaviors of sequences and functions in both mathematics and real-world applications.

Exponential Functions

Exponential functions are those in which a constant base is raised to a variable exponent, such as \( 2^n \). These functions are particularly known for their rapid growth rate, which becomes evident when analyzing sequences.
For our function \( 2^n \), as \( n \) increases, the value of \( 2^n \) increases exponentially. This particular exponential growth implies that \( 2^n \) will quickly become much larger than \( n \) itself. The exponential function surpasses any polynomial function for sufficiently large values of \( n \). This is the key reason why \( \lim_{n \to \infty} 2^n = \infty \).
Exponential functions like \( 2^n \) illustrate not just mathematical concepts but also real-world phenomena like population growth, radioactive decay, and interest calculations. Understanding their behavior is vital for predicting trends and occurrences in various scientific fields.

Proof by Induction

Proof by induction is a powerful mathematical technique used to prove statements that are assumed to be true for every natural number. This involves two major steps: proving a base case and showing the truth of one step implies the truth of the next.
In this exercise, we first verified that the base case \( S_1 \) is true: indeed, \( 1 < 2^1 \). Next, we assumed that \( S_k \) is true for some arbitrary number \( k \), which is our inductive hypothesis. We then showed that if \( S_k \) is true, then \( S_{k+1} \) must also be true. Through these steps, we concluded that since the base case holds and the implication from \( S_k \) to \( S_{k+1} \) is also true, all statements \( S_n \) must be true for every natural number \( n \).
Proof by induction is particularly useful in mathematical analysis because it allows mathematicians to establish the truth of infinitely many cases with finite effort. In our context, it confirms the rapid growth of \( 2^n \) compared to \( n \). This was crucial for showing that \( 2^n \) becomes very large as \( n \) increases, effectively validating \( \lim_{n \to \infty} 2^n = \infty \).