Problem 41
Question
Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(\frac{n+1}{2 n}\right)\left(1-\frac{1}{n}\right) $$
Step-by-Step Solution
Verified Answer
The sequence converges with a limit of \( \frac{1}{2} \).
1Step 1: Simplify the Sequence Expression
We begin by simplifying the expression for the sequence: \[ a_n = \left( \frac{n+1}{2n} \right) \left( 1 - \frac{1}{n} \right) \]Start by simplifying each part separately:First, \( \frac{n+1}{2n} \) can be rewritten as \( \frac{n}{2n} + \frac{1}{2n} = \frac{1}{2} + \frac{1}{2n} \).Second, \( 1 - \frac{1}{n} \) remains as is. Multiplying these simplified expressions, we have:\[ a_n = \left( \frac{1}{2} + \frac{1}{2n} \right) \left( 1 - \frac{1}{n} \right) \]
2Step 2: Distribute and Simplify Further
Now, distribute the multiplication:\[ a_n = \left( \frac{1}{2} \right) \left( 1 - \frac{1}{n} \right) + \left( \frac{1}{2n} \right) \left( 1 - \frac{1}{n} \right) \]Expanding each term gives:\[ a_n = \frac{1}{2} - \frac{1}{2n} + \frac{1}{2n} - \frac{1}{2n^2} \]Notice that the \( \frac{1}{2n} \) terms cancel each other, simplifying to:\[ a_n = \frac{1}{2} - \frac{1}{2n^2} \]
3Step 3: Determine the Limit as n Approaches Infinity
To determine the limit of \( a_n \) as \( n \to \infty \), examine the expression:\[ \lim_{{n \to \infty}} a_n = \lim_{{n \to \infty}} \left( \frac{1}{2} - \frac{1}{2n^2} \right) \]As \( n \to \infty \), the term \( \frac{1}{2n^2} \) approaches zero, thus:\[ \lim_{{n \to \infty}} a_n = \frac{1}{2} - 0 = \frac{1}{2} \]
4Step 4: Conclusion on Convergence
Since the limit \( \frac{1}{2} \) exists and is a finite number, the sequence \( a_n \) is convergent. The limit of this convergent sequence is \( \frac{1}{2} \).
Key Concepts
Limit of a SequenceInfinity in SequencesSimplification of Expressions
Limit of a Sequence
In mathematics, the term "limit of a sequence" refers to the value that the terms of a sequence approach as the index number ( extit{n}) goes to infinity. A sequence has a limit if, as we consider each successive term, they get "closer" to a certain number. This concept is fundamental in calculus and real analysis.
For example, in our sequence defined as \(a_{n}=(\frac{n+1}{2n})(1-\frac{1}{n})\). After simplification and rearrangement, the sequence becomes \(a_{n}=\frac{1}{2} - \frac{1}{2n^2}\).
As \(n\) becomes very large, the term \(\frac{1}{2n^2}\) becomes very small, eventually approaching zero. Consequently, we see that:
For example, in our sequence defined as \(a_{n}=(\frac{n+1}{2n})(1-\frac{1}{n})\). After simplification and rearrangement, the sequence becomes \(a_{n}=\frac{1}{2} - \frac{1}{2n^2}\).
As \(n\) becomes very large, the term \(\frac{1}{2n^2}\) becomes very small, eventually approaching zero. Consequently, we see that:
- \a_{n} approaches \(\frac{1}{2}\).
- This sequence's limit is \(\frac{1}{2}\).
- We say the sequence converges to this value.
Infinity in Sequences
Infinity plays a significant role when analyzing sequences. When discussing infinity in sequences, it's important to understand that it describes the behavior of the sequence as \(n\) becomes very large, beyond any finite limit.
In the context of the sequence \(a_{n}=\frac{1}{2} - \frac{1}{2n^2}\) derived from the formula \(a_{n}=(\frac{n+1}{2n})(1-\frac{1}{n})\), we consider what happens as \(n\) approaches infinity.
Here's what happens as \(n\) increases:
- The portion \(\frac{1}{2n^2}\) becomes negligible.- We see \(a_{n}\) approaches a constant value, specifically \(\frac{1}{2}\).- This shows that the sequence shows stability or convergence.
It's crucial in calculus and mathematical analysis to understand how terms behave when sequences extend towards infinity.
In the context of the sequence \(a_{n}=\frac{1}{2} - \frac{1}{2n^2}\) derived from the formula \(a_{n}=(\frac{n+1}{2n})(1-\frac{1}{n})\), we consider what happens as \(n\) approaches infinity.
Here's what happens as \(n\) increases:
- The portion \(\frac{1}{2n^2}\) becomes negligible.- We see \(a_{n}\) approaches a constant value, specifically \(\frac{1}{2}\).- This shows that the sequence shows stability or convergence.
It's crucial in calculus and mathematical analysis to understand how terms behave when sequences extend towards infinity.
Simplification of Expressions
Simplifying expressions is a key step in understanding the behavior of sequences. It allows us to see clearer patterns and relationships, especially when dealing with limits and convergence.
In our sequence \(a_{n}=(\frac{n+1}{2n})(1-\frac{1}{n})\), simplification was performed in steps:
Through simplification:
In our sequence \(a_{n}=(\frac{n+1}{2n})(1-\frac{1}{n})\), simplification was performed in steps:
- First, the term \(\frac{n+1}{2n}\) was simplified to \(\frac{1}{2} + \frac{1}{2n}\).
- Then, expanding \(\left(\frac{1}{2} + \frac{1}{2n}\right)(1-\frac{1}{n})\) allowed us to cancel terms and simplify directly to \(\frac{1}{2} - \frac{1}{2n^2}\).
Through simplification:
- We recognize convergent patterns.
- Understand the importance of each part of the expression.
Other exercises in this chapter
Problem 41
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