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Magazine Chapter 11: Problem 52
Determine whether the sequence converges or diverges. If it converges, find the limit. $$ a_{n}=n-\sqrt{n+1} \sqrt{n+3} $$
Short Answer
Expert verified
The sequence converges, limit is \(-2\).
Step by step solution
01
Analyze the Sequence Expression
We need to determine if \( a_n = n - \sqrt{n+1} \sqrt{n+3} \) converges or diverges. First, let's analyze the structure of the sequence expression. As \( n \to \infty \), both \( \sqrt{n+1} \) and \( \sqrt{n+3} \) behave like \( \sqrt{n} \). Therefore, \( \sqrt{n+1} \sqrt{n+3} \approx \sqrt{n} \cdot \sqrt{n} = n \).
02
Expand the Expression at Infinity
Perform algebraic manipulations for larger \( n \):\[\sqrt{n+1}\sqrt{n+3} = \sqrt{(n+1)(n+3)} = \sqrt{n^2 + 4n + 3}\]Thus, the sequence becomes:\[a_n = n - \sqrt{n^2 + 4n + 3}\]
03
Simplify Using Binomial Approximation
Apply the expansion for square roots:\[\sqrt{n^2 + 4n + 3} \approx n + \frac{4n + 3}{2n} = n + 2 + \frac{3}{2n}\]Thus:\[a_n = n - (n + 2 + \frac{3}{2n}) = -2 - \frac{3}{2n}\]
04
Evaluate the Limit
Evaluate the limit as \( n \to \infty \):\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(-2 - \frac{3}{2n}\right)\]The term \( \frac{3}{2n} \to 0 \) as \( n \to \infty \), so the limit is:\[-2\]
05
Conclusion on Convergence
Since the limit exists and equals \(-2\), the sequence \( a_n = n - \sqrt{n+1} \sqrt{n+3} \) converges. The limit is \(-2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Sequence
When we talk about the limit of a sequence, we ask whether the sequence approaches a fixed value as the term number, usually denoted by \( n \), becomes very large. If a sequence has a limit, it is said to converge to that limit. On the other hand, if no such value exists, the sequence diverges.
In our exercise, we have the sequence given by \( a_n = n - \sqrt{n+1}\sqrt{n+3} \). To find out if this sequence has a limit, we examined how the terms behave as \( n \to \infty \). Through our calculation, we determined that the limit of the sequence \( a_n \) is \(-2\).
### Key Principles
In our exercise, we have the sequence given by \( a_n = n - \sqrt{n+1}\sqrt{n+3} \). To find out if this sequence has a limit, we examined how the terms behave as \( n \to \infty \). Through our calculation, we determined that the limit of the sequence \( a_n \) is \(-2\).
### Key Principles
- For the sequence to converge, its terms must approach a single fixed number as \( n \) increases.
- Algebraic simplification and careful analysis around behavior at infinity are often used to determine limits.
- In the example, we simplified the sequence using various mathematical approximations, such as binomial approximation, to deduce its limit.
Binomial Approximation
The binomial approximation is a powerful tool when dealing with roots and polynomial expansion. It helps us simplify expressions, especially when considering terms as \( n \to \infty \). In this exercise, the square root term \( \sqrt{n^2 + 4n + 3} \) was expanded using a form of binomial approximation.
### How Approximations Work
Using the approximation, we expanded the square root function around a large \( n \). The approximation was:\[\sqrt{n^2 + 4n + 3} \approx n + \frac{4n + 3}{2n} = n + 2 + \frac{3}{2n}\] This technique simplifies the analysis because:
### How Approximations Work
Using the approximation, we expanded the square root function around a large \( n \). The approximation was:\[\sqrt{n^2 + 4n + 3} \approx n + \frac{4n + 3}{2n} = n + 2 + \frac{3}{2n}\] This technique simplifies the analysis because:
- We can drop smaller term contributions when compared to dominant \( n \) terms.
- It helps us identify the leading behavior of sequences or series as \( n \) becomes very large.
- Binomial approximation is useful for determining limits, as higher-order terms become negligible.
Algebraic Manipulation
Algebraic manipulation plays a crucial role in finding simplified forms of sequences and functions. It involves using algebraic rules to rewrite and reduce expressions, making them easier to analyze, especially when dealing with sequences.
### Steps of Manipulation in This Context
For the sequence \( a_n = n - \sqrt{n+1} \sqrt{n+3} \), algebraic manipulation was essential:
### Steps of Manipulation in This Context
For the sequence \( a_n = n - \sqrt{n+1} \sqrt{n+3} \), algebraic manipulation was essential:
- First, we recognized that \( \sqrt{n+1} \sqrt{n+3} \) could be rewritten as \( \sqrt{(n+1)(n+3)} = \sqrt{n^2 + 4n + 3} \).
- Then, by using algebraic simplification alongside binomial approximation, the expression was further simplified to \( a_n = -2 - \frac{3}{2n} \).
- The algebraic manipulation helped isolate terms and identify the sequence's behavior as \( n \to \infty \).
