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Chapter 16: Problem 29

Prove by mathematical induction that \(1 / n ! \leq 1 / 2^{n-1}\).

Short Answer

Expert verified
The inequality \( \frac{1}{n!} \leq \frac{1}{2^{n-1}} \) is true for all positive integers n by mathematical induction.

Step by step solution

01

- Base Case

Verify the base case for the smallest possible value of n, which is typically n = 1. Calculate both sides of the inequality: Left Side: \( \frac{1}{1!} = 1 \)Right Side: \( \frac{1}{2^{1-1}} = \frac{1}{2^0} = 1\)Since \( 1 = 1 \), the base case holds.
02

- Inductive Hypothesis

Assume the inequality is true for some arbitrary positive integer k: Assume \( \frac{1}{k!} \leq \frac{1}{2^{k-1}} \). This is the inductive hypothesis.
03

- Inductive Step

Show that if the inequality holds for n = k, it also holds for n = k + 1. We need to prove: \( \frac{1}{(k+1)!} \leq \frac{1}{2^k} \)Start with the left side: \( \frac{1}{(k+1)!} = \frac{1}{(k+1) \cdot k!} \)According to the inductive hypothesis: \( \frac{1}{k!} \leq \frac{1}{2^{k-1}} \)So: \( \frac{1}{(k+1) \cdot k!} \leq \frac{1}{(k+1) \cdot 2^{k-1}} \)Simplify the right side: \( \frac{1}{(k+1) \cdot 2^{k-1}} = \frac{1}{2^k} \)Therefore: \( \frac{1}{(k+1)!} \leq \frac{1}{2^k} \).
04

- Conclusion

Since both the base case and inductive step have been verified, by mathematical induction, the statement \( \frac{1}{n!} \leq \frac{1}{2^{n-1}} \) is true for all positive integers n.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mathematical induction
Mathematical induction is a powerful technique used to prove statements about integers. The process involves two main steps: the base case and the inductive step. By first proving the statement for an initial value (usually the smallest integer, like 1), and then proving that if the statement holds for an arbitrary integer, it holds for the next integer as well, we can conclude that the statement is true for all integers.
factorial inequality
A factorial inequality refers to comparing expressions that involve factorials. Factorials are products of all positive integers up to a certain number (denoted as n!). In this problem, we need to prove that \( \frac{1}{n!} \leq \frac{1}{2^{n-1}} \). The challenge is to establish this relation for all positive integers n using mathematical induction. Factorial expressions often grow quickly, so inequalities involving them can be tricky and require careful manipulation.
inductive hypothesis
The inductive hypothesis is the assumption that a given statement is true for some arbitrary positive integer k. For our problem, we assume \( \frac{1}{k!} \leq \frac{1}{2^{k-1}} \) to be true. This forms the basis we use to prove that the inequality holds for k+1. The hypothesis is crucial as it forms the bridge from one integer to the next in the chain of reasoning that drives mathematical induction forward.
base case
The base case is the first step in a proof by mathematical induction. It involves verifying that the statement in question holds true for the smallest possible value, often n = 1. For this problem, the base case requires evaluating \( \frac{1}{1!} = 1 \) and \ \frac{1}{2^{1-1}} = 1 \. Since both sides of the inequality are equal (1 = 1), the base case is satisfied. This initial proof anchors the induction process, allowing us to use the inductive hypothesis to then show the statement holds for all natural numbers.

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Most popular questions from this chapter

Assume that the constant 0 has a power-series representation \(\sum_{n=0}^{+\infty} c_{\mu} x^{n}\), where the radius of convergence \(R>0 .\) Prove that \(c_{n}=0\) for all \(n\)

Find the error if the sum of the first four terms is used as an approximation to the sum of the given infinite series.\(\sum_{n=1}^{+\infty}(-1)^{n} \frac{1}{n !}\)

Determine \(a_{n}(n=0,1,2,3,4)\) so that the polynomial $$ f(x)=3 x^{4}-17 x^{3}+35 x^{2}-32 x+17 $$ is written in the form. $$ f(x)=a_{4}(x-1)^{4}+a_{3}(x-1)^{3}+a_{2}(x-1)^{2}+a_{1}(x-1)+a_{0} $$

A function \(f\) is defined by a power series. In each exercise do the following: (a) Find the radius of convergence of the given power series and the domain of \(f\); (b) write the power series which defines the function \(f^{\prime}\) and find its radius of convergence by using methods of Sec. \(16.7\) (thus verifying Theorem 16.8.1); (c) find the domain of \(f^{\prime}\).\(f(x)=\sum_{n=1}^{+\infty} \frac{x^{n}}{\sqrt{n}}\)

Compute the value of the definite integral accurate to four decimal places.\(\int_{0}^{1 / 2} \sqrt{1-x^{3}} d x\)
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