Problem 127
Question
The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\} .\)
Step-by-Step Solution
Verified Answer
The sequence \( \{ \frac{n}{n+1} \} \) exceeds any value less than 1, confirming 1 as the least upper bound.
1Step 1: Understand the Sequence
The sequence given is \( \{ \frac{n}{n+1} \} \), where \( n \) is a natural number. Each term of the sequence can be expressed as \( \frac{n}{n+1} = 1 - \frac{1}{n+1} \). This shows that as \( n \) becomes very large, \( \frac{1}{n+1} \) becomes very small, making the term approach 1.
2Step 2: Explore Condition for M
We need to show that for any \( M < 1 \), you can find an \( N \) such that \( \frac{n}{n+1} > M \) for all \( n > N \). Since \( \frac{n}{n+1} = 1 - \frac{1}{n+1} \), this becomes \( 1 - \frac{1}{n+1} > M \). Rearranging terms gives us \( \frac{1}{n+1} < 1 - M \).
3Step 3: Solve for N
The inequality \( \frac{1}{n+1} < 1 - M \) implies that \( n+1 > \frac{1}{1-M} \). Thus, \( n > \frac{1}{1-M} - 1 \). If we let \( N = \lceil \frac{1}{1-M} - 1 \rceil \), then for any \( n > N \), the inequality will be satisfied.
4Step 4: Conclude the Upper Bound
Since \( n /(n+1) < 1 \) for every \( n \) and as shown, it exceeds any number \( M < 1 \) eventually as \( n \) increases, this means that 1 is indeed the least upper bound for the sequence \( \{ \frac{n}{n+1} \} \).
Key Concepts
Sequence AnalysisConvergenceInequalities
Sequence Analysis
Understanding the sequence \( \left\{ \frac{n}{n+1} \right\} \) is crucial for analyzing its behavior as \( n \) increases. This sequence is defined as each term being \( \frac{n}{n+1} \), where \( n \) is a natural number. A natural number is a positive integer starting from 1, so examples of such numbers include 1, 2, 3, and so on. By reworking the expression, each term can be seen as \( \frac{n}{n+1} = 1 - \frac{1}{n+1} \). This is a significant simplification because it shows how the term evolves: as \( n \) increases, \( \frac{1}{n+1} \) diminishes, meaning the sequence moves closer and closer to 1.
This simplification helps us understand the sequence's behavior without computing numerous terms. Knowing the expression \( 1 - \frac{1}{n+1} \) highlights its tendency to approach a particular value, forming the foundation for further analysis in terms of bounds and convergence.
This simplification helps us understand the sequence's behavior without computing numerous terms. Knowing the expression \( 1 - \frac{1}{n+1} \) highlights its tendency to approach a particular value, forming the foundation for further analysis in terms of bounds and convergence.
Convergence
Convergence is all about determining where a sequence is headed as its terms progress. With the sequence \( \left\{ \frac{n}{n+1} \right\} \), convergence implies finding the value the sequence terms approach as \( n \) becomes very large.
By observing the expression \( 1 - \frac{1}{n+1} \), we note that \( \frac{1}{n+1} \) shrinks towards zero as \( n \) grows larger. Consequently, each term of the sequence approaches 1. In formal terms, we say that the sequence \( \left\{ \frac{n}{n+1} \right\} \) converges to 1.
By observing the expression \( 1 - \frac{1}{n+1} \), we note that \( \frac{1}{n+1} \) shrinks towards zero as \( n \) grows larger. Consequently, each term of the sequence approaches 1. In formal terms, we say that the sequence \( \left\{ \frac{n}{n+1} \right\} \) converges to 1.
- This outcome is significant because convergence informs us about the behavior of the entire sequence.
- It helps analyze whether a sequence reaches a certain limit.
Inequalities
Inequalities play a vital role in proving the least upper bound of a sequence. For our sequence \( \left\{ \frac{n}{n+1} \right\} \), the concept of inequalities comes into play when we need to show that for any number \( M < 1 \), we can find a term in the sequence larger than \( M \).
Let's break it down: given that \( \frac{n}{n+1} = 1 - \frac{1}{n+1} \), showing that each term exceeds \( M \) when \( M < 1 \) can be translated to finding an \( N \) such that \( \frac{n}{n+1} > M \) for all \( n > N \).
Let's break it down: given that \( \frac{n}{n+1} = 1 - \frac{1}{n+1} \), showing that each term exceeds \( M \) when \( M < 1 \) can be translated to finding an \( N \) such that \( \frac{n}{n+1} > M \) for all \( n > N \).
- This inequality is rearranged to state \( \frac{1}{n+1} < 1 - M \), guiding us to solve for \( N \) as \( n > \frac{1}{1-M} - 1 \).
- Setting \( N = \left\lceil \frac{1}{1-M} - 1 \right\rceil \), guarantees that for any \( n > N \), the terms satisfy \( \frac{n}{n+1} > M \).
- This ensures that 1 remains the sequence’s least upper bound, as each term eventually surpasses any \( M < 1 \).
Other exercises in this chapter
Problem 125
In Exercises \(125-126,\) use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0$$
View solution Problem 126
In Exercises \(125-126,\) use the definition of convergence to prove the given limit. $$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n^{2}}\right)=1$$
View solution Problem 128
Uniqueness of least upper bounds Show that if \(M_{1}\) and \(M_{2}\) are least upper bounds for the sequence \(\left\\{a_{n}\right\\},\) then \(M_{1}=M_{2}\) T
View solution Problem 130
Prove that if \(\left\\{a_{n}\right\\}\) is a convergent sequence, then to every positive number \(\epsilon\) there corresponds an integer \(N\) such that for a
View solution