Problem 1
Question
Match the formulas with the descriptions of the behavior of the sequence as \(n \rightarrow \infty\). 1\. \(s_{n}=n(n+1)-1\) 2\. \(s_{n}=1 /(n+1)\) 3\. \(s_{n}=3-1 / n\) 4\. \(s_{n}=n \sin (n) /(n+1)\) 5\. \(s_{n}=(n+1) / n\) A. does not converge, but doesn'\operatorname{tg} o ~ t o ~ \(\pm \infty\) B. converges to three from below C. diverges to \(\infty\) D. converges to one from above E. converges to zero through positive numbers
Step-by-Step Solution
Verified Answer
1. C, 2. E, 3. B, 4. A, 5. D.
1Step 1: Match the first formula
Consider the formula for sequence 1: \[ s_{n} = n(n+1) - 1 \] As \( n \rightarrow \infty \), the term \( n(n+1) \) grows quadratically, hence \[ s_{n} \rightarrow \infty \] Therefore, the sequence diverges to infinity. The matching option is C.
2Step 2: Match the second formula
Consider the formula for sequence 2: \[ s_{n} = \frac{1}{n+1} \] As \( n \rightarrow \infty \), the denominator grows large, making the overall fraction approach zero: \[ s_{n} \rightarrow 0^{+} \] The sequence converges to zero through positive numbers. The matching option is E.
3Step 3: Match the third formula
Consider the formula for sequence 3: \[ s_{n} = 3 - \frac{1}{n} \] As \( n \rightarrow \infty \), \( \frac{1}{n} \) approaches zero, making \[ s_{n} \rightarrow 3 \] Since \( s_{n} \) approaches 3 from below, the matching option is B.
4Step 4: Match the fourth formula
Consider the formula for sequence 4: \[ s_{n}=\frac{n \sin(n)}{n+1} \] Given that \( n \sin(n) \) oscillates between \( -n \) and \( n \), the sequence does not settle into a single value but also does not diverge to \( \infty \). Therefore, it does not converge but also doesn’t go to \( \pm \infty \). The matching option is A.
5Step 5: Match the fifth formula
Consider the formula for sequence 5: \[ s_{n} = \frac{n+1}{n} \] As \( n \rightarrow \infty \), \[ s_{n} = 1 + \frac{1}{n} \] and \( \frac{1}{n} \) approaches zero, so \[ s_{n} \rightarrow 1^{+} \] The sequence converges to one from above. The matching option is D.
Key Concepts
ConvergenceDivergenceLimits
Convergence
When we talk about convergence in sequences, we refer to how the terms of the sequence behave as the index (usually denoted as 'n') approaches infinity. A sequence converges if its terms get closer and closer to a specific value, known as the limit, as n gets larger. This specific value is where the sequence 'settles'.
Some key points on convergence:
Some key points on convergence:
- A sequence \( s_n \) converges to a limit L if for every small positive number \( \varepsilon \), there is a corresponding large number N such that for all n \> N, \( |s_n - L| < \varepsilon \).
- Sequences can converge from above or below, as seen in examples like \( s_n = 3 - \frac{1}{n} \) converging to 3 from below and \( s_n = \frac{n+1}{n} \) converging to 1 from above.
Divergence
Divergence in sequences is, in many ways, the opposite of convergence. A sequence is said to diverge if its terms do not approach a specific limit as the index n approaches infinity. Instead of getting closer to a particular value, the terms either increase or decrease without bound, or they oscillate without settling on a particular value.
Key observations about divergence include:
Key observations about divergence include:
- If a sequence grows indefinitely, it diverges to \( \infty \) or \( -\infty \). For example, \( s_n = n(n+1) - 1 \) diverges to infinity.
- Some sequences do not settle into any particular value nor do they grow without bound; they simply oscillate. An example is \( s_n = \frac{n \sin(n)}{n+1} \), which neither converges nor diverges to infinity.
Limits
The limit of a sequence defines the value that the terms of the sequence approach as the index n tends to infinity. Limits are a fundamental concept in calculus and are vital in understanding the long-term behavior of sequences.
Important aspects of limits include:
Important aspects of limits include:
- Formally, a sequence \( s_n \) has a limit L if for every \( \varepsilon > 0 \), there exists an integer N such that for all \( n \geq N \, |s_n - L| < \varepsilon \).
- In the sequence \( s_n = \frac{1}{n+1} \), as n approaches infinity, \( s_n \) approaches 0. Therefore, the limit of this sequence is 0.
- Limits can be finite, like the limit of \( s_n = 3 - \frac{1}{n} \), which is 3, or infinite, where the sequence grows without bounds.
Other exercises in this chapter
Problem 1
Find the Taylor polynomials of degree \(n\) approximating \(\cos (3 x)\) for \(x\) near 0 : For \(n=2, P_{2}(x)=\) For \(n=4, P_{4}(x)=\) For \(n=6, P_{6}(x)=\)
View solution Problem 1
Given: \(A_{n}=\frac{80}{8^{n}}\) Determine: (a) whether \(\sum_{n=1}^{\infty}\left(A_{n}\right)\) is convergent. (b) whether \(\left\\{A_{n}\right\\}\) is conv
View solution Problem 2
For the following alternating series, \(\sum_{n=1}^{\infty} a_{n}=0.45-\frac{(0.45)^{3}}{3 !}+\frac{(0.45)^{5}}{5 !}-\frac{(0.45)^{7}}{7 !}+\ldots\) how many te
View solution Problem 2
Consider the series \(\sum_{n=1}^{\infty} \frac{10}{n+2}\). Let \(s_{n}\) be the \(\mathrm{n}\) -th partial sum; that is, $$ s_{n}=\sum_{i=1}^{n} \frac{10}{i+2}
View solution