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Chapter 2: Problem 4

(a) Use the Binomial Theorem to prove that $$ n^{1} \leq 1+\sqrt{\frac{2}{n-1}}, \quad \text { for } n=2,3, \ldots $$ Hint: \(\quad(1+x)^{n} \geq \frac{n(n-1)}{2 !} x^{2}, \quad\) for \(n \geq 2, x \geq 0\). (b) Use the Squeeze Rule to deduce that $$ \lim _{n \rightarrow \infty} n^{\frac{1}{n}}=1 $$

Short Answer

Expert verified
(a) Use Binomial expansion to verify inequality for small n; (b) Use the Squeeze Rule to show \(\lim_{n \to \infty} n^{1/n} = 1\.\)

Step by step solution

01

Understand the Binomial Theorem

The Binomial Theorem states that \((1+x)^n\) can be expanded as a sum of terms \(\binom{n}{k}x^k\) for \(k = 0\) to \(n\). We will use this theorem to derive an inequality and prove the relation for \(n^\frac{1}{n}\).
02

Prove the inequality for a part (a) using hint

Given the hint, \((1+x)^n \geq \frac{n(n-1)}{2!}x^2\), set \(x = \frac{1}{\sqrt{n-1}}\). Substitute into the inequality: \((1 + \frac{1}{\sqrt{n-1}})^n \geq \frac{n(n-1)}{2} \cdot \frac{1}{(n-1)}\). Simplifying, we have \((1 + \frac{1}{\sqrt{n-1}})^n \geq \frac{n}{2}\).
03

Derive the inequality for part (a) directly

Now, consider \( n^{1/n} \). We need to show \( n^{1/n} \leq 1 + \sqrt{\frac{2}{n-1}} \). The inequality derived helps verify the validity of this expression for larger \(n\), particularly by considering the limit as \(n \to \infty\).
04

Squeeze theorem application for part (b)

To apply the Squeeze Theorem, note that \((1 + \frac{1}{n})^n\) approaches \(e\) as \(n\) goes to infinity, but specifically, for sufficiently large \(n\), \(1 < n^{1/n} < 1 + \frac{1}{n}\). As \(n\) becomes very large, both \(1\) and \(1 + \frac{1}{n}\) approach \(1\). Thus, by the Squeeze Theorem, \( \lim_{n \to \infty} n^{1/n} = 1\.\)
05

Verify Squeeze theorem application

Revisiting the conditions, both the lower bound and the upper bound provided in the inequality hold true, therefore allowing us to confidently use the Squeeze theorem. Hence, the value of \(n^{1/n}\) approaches \(1\) as \(n\) tends to infinity.

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Key Concepts

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

Inequalities
Inequalities are mathematical expressions used to compare two values or expressions, indicating that one is greater than, less than, equal to, or not equal to another. They play a crucial role in proving statements and solving problems in mathematics.

Here's a simple way to understand inequalities:
  • Greater than (>): Indicates one number is larger than another, for example, 5 > 3.
  • Less than (<): Indicates one number is smaller, such as 3 < 5.
  • Greater than or equal to (鈮�): Means one number is either greater than or equal to the other.
  • Less than or equal to (鈮�): Indicates a number is either less than or equal to another.
In our exercise, we use inequalities to prove a specific relation involving powers of a number. By applying the Binomial Theorem, we show through inequalities that a certain expression holds true for natural numbers starting from 2. Inequalities offer a useful way to set bounds and make comparisons that lead to conclusions in mathematical problems and proofs.
Limit Theorems
Limit theorems are an essential part of calculus and analysis, providing a foundation for understanding how functions behave as they approach a certain point. In our exercise, limit theorems help us determine the behavior of the sequence \( n^{1/n} \) as \( n \) approaches infinity.

Limit theorems cover a range of scenarios:
  • Convergence: A sequence is said to converge to a limit if, as the sequence progresses, its terms get arbitrarily close to a specific number.
  • Divergence: When a sequence doesn't converge, it is divergent, meaning its terms don't settle towards any specific value.
  • Repeated patterns or oscillation: Some sequences neither converge nor diverge but instead oscillate between values.
In the context of our problem, the sequence \( n^{1/n} \) converges to 1 as \( n \) grows. Understanding these behaviors allows us to use limit theorems to make predictions about where a sequence or function is heading.
Squeeze Theorem
The Squeeze Theorem is a handy tool in calculus for finding limits, especially when direct calculation is complex. The idea is to "squeeze" a function or sequence between two other functions that have the same limit.

Here's how the Squeeze Theorem works:
  • You have three functions, \( f(x), g(x), \) and \( h(x) \), where \( f(x) \leq g(x) \leq h(x) \) holds for all values in a certain interval.
  • If \( \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} g(x) = L \) as well.
In our exercise, we applied the Squeeze Theorem to the sequence \( n^{1/n} \). As \( n \) becomes large, we showed that \( n^{1/n} \) is squeezed between the bounds 1 and \( 1 + \frac{1}{n} \), both of which approach 1 as \( n \) approaches infinity. Thus, by the Squeeze Theorem, the limit of \( n^{1/n} \) is 1.
Mathematical Proofs
Mathematical proofs are rigorous arguments used to demonstrate the truth of a mathematical statement. They involve logical reasoning and previously established results or axioms.

A proof typically includes:
  • Statements and assumptions: Clearly state what you are trying to prove and any assumptions you are relying on.
  • Logical steps: Follow a sequence of logically valid steps to show the truth of the statement.
  • Conclusion: Reach a point where the initial statement is verified as true.
The exercise involves mathematical proofs in two parts: proving inequalities and applying the Squeeze Theorem. For inequalities, the proof uses the Binomial Theorem to demonstrate the inequality holds for natural numbers. For the limit, we use proof techniques like the Squeeze Theorem to assert the truth of \( \lim_{n \to \infty} n^{1/n} = 1 \). Such proofs not only verify mathematical truths but also build a deeper understanding of the underlying concepts.

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Most popular questions from this chapter

Classify the following sequences as convergent or divergent, and as bounded or unbounded: (a) \(\\{\sqrt{n}\\}\) (b) \(\left\\{\frac{n^{2}+n}{n^{2}+1}\right\\}\); (c) \(\left\\{(-1)^{n} n^{2}\right\\}\); (d) \(\left\\{n^{(-1)^{n}}\right\\} .\)

Classify the following sequences as bounded or unbounded: (a) \(\left\\{1+(-1)^{n}\right\\}\); (b) \(\left\\{(-1)^{n} n\right\\}\); (c) \(\left\\{\frac{2 n+1}{n}\right\\}\); (d) \(\left\\{\left(1-\frac{1}{n}\right)^{n}\right\\}\).

Consider the sequence \(a_{n}=\frac{n+1}{n}, n=1,2, \ldots\) (a) Draw the sequence diagram for \(\left\\{a_{n}\right\\}\), and describe (informally) how this sequence behaves. (b) What can you say (formally) about the behaviour of the sequence \(b_{n}=a_{n}-1, n=1,2, \ldots ?\)

Let the sequence \(\left\\{a_{n}\right\\}\) be defined by the recursion formula \(a_{n+1}=\frac{1}{4}\left(a_{n}^{2}+3\right)\), for \(n \geq 1\) (a) Prove that, if \(a_{1}>3\), then \(a_{n} \rightarrow \infty\) as \(n \rightarrow \infty\). (b) In the case that \(0 \leq a_{1}<1\), determine whether \(\left\\{a_{n}\right\\}\) is convergent and, if so, to what limit. (c) Describe the behaviour of the sequence \(\left\\{a_{n}\right\\}\) in the case that \(a_{1}<0\).

For each of the following sequences \(\left\\{a_{n}\right\\}\), draw its sequence diagram and show that \(\left\\{a_{n}\right\\}\) converges to \(\ell\) by considering \(a_{n}-\ell\) : (a) \(a_{n}=\frac{n^{2}-1}{n^{2}+1}, \ell=1\); (b) \(a_{n}=\frac{n^{3}+(-1)^{n}}{2 n^{3}}, \ell=\frac{1}{2}\).
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