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Chapter 11: Problem 9

Determine the boundedness and monotonicity of the sequence with \(a_{n},\) as indicated. $$\frac{2}{n}$$.

Short Answer

Expert verified
The sequence \(a_n = \frac{2}{n}\) is monotonically decreasing and bounded, with an upper limit of 2 and a lower limit of 0.

Step by step solution

01

Check if the sequence is monotonic

To determine if the sequence is monotonic, we should analyze the ratio or difference between consecutive terms: \(a_{n+1}\) and \(a_n\). Let's find the difference between consecutive terms: \[ a_{n + 1} - a_n = \frac{2}{n + 1} - \frac{2}{n}. \]
02

Find a common denominator and simplify

Now, let's find a common denominator and simplify the difference we found in Step 1: \[ \frac{2}{n + 1} - \frac{2}{n} = \frac{2n - 2(n + 1)}{n(n + 1)} = \frac{-2}{n(n + 1)}. \]
03

Determine monotonicity

From Step 2, we have found that the difference between consecutive terms is \(\frac{-2}{n(n + 1)}\). Since the numerator is negative and both denominators are positive, the difference is negative for all n. This means that the terms of the sequence are always decreasing. Therefore, the sequence is monotonically decreasing.
04

Determine if the sequence is bounded

In order to determine if the sequence is bounded, we need to show that the terms of the sequence are confined within an upper limit and a lower limit. Let's analyze the terms of the sequence when n approaches infinity: \[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{2}{n} = 0. \] As n gets larger, the terms of the sequence \(a_n = \frac{2}{n}\) approach 0. So, the lower limit for the sequence is 0. Now, let's analyze the upper limit: For all \(n \geq 1\), we have \(a_n = \frac{2}{n} \leq \frac{2}{1}\), which means that the terms of the sequence are always less than or equal to 2. So, the upper limit for the sequence is 2. Since the sequence has an upper limit of 2 and a lower limit of 0, the sequence is bounded. To summarize, The sequence \(a_n = \frac{2}{n}\) is monotonic and decreasing, and it is also bounded with an upper limit of 2 and a lower limit of 0.

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Key Concepts

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

Monotonic Sequences
A sequence is considered monotonic if its terms are either entirely non-increasing or non-decreasing throughout its span. In simple terms, this means the sequence either always goes up or always goes down as you move from one term to the next. There are two main types of monotonic sequences:
  • Monotonically Increasing: Each term is greater than or equal to the previous one.
  • Monotonically Decreasing: Each term is less than or equal to the previous one.
To check if a specific sequence is monotonic, we often analyze the difference or ratio between consecutive terms.
For example, in the sequence \( a_n = \frac{2}{n} \), by calculating the difference between consecutive terms, we found \( a_{n+1} - a_n = \frac{-2}{n(n+1)} \), which is negative for all \( n \). This indicates that the sequence is monotonically decreasing, meaning every term is smaller than the one before it.
Bounded Sequences
A sequence is said to be bounded if its terms do not exceed certain fixed limits, both above and below. Essentially, it means the sequence stays within a certain range.
  • Upper Bound: There is a number that every term in the sequence is less than or equal to.
  • Lower Bound: There is a number that every term in the sequence is greater than or equal to.
In our example of \( a_n = \frac{2}{n} \), as \( n \) becomes very large, the terms \( \frac{2}{n} \) get closer and closer to 0, showing a lower bound of 0.
For the upper bound, since \( \frac{2}{n} \leq 2 \) for all \( n \geq 1 \), the sequence is confined above by the fraction \( 2 \). Thus, the sequence is bounded with the boundaries set between 0 and 2.
Sequence Limits
The limit of a sequence is the value that the terms of a sequence approach as the index (often denoted \( n \)) goes to infinity. It provides a way to understand the long-term behavior of the sequence.For the sequence \( a_n = \frac{2}{n} \), as \( n \) increases indefinitely, the terms get closer and closer to 0. We express this formally using limits: \[\lim_{n\to\infty} \frac{2}{n} = 0.\]This means the limit of the sequence is 0, giving us crucial information about where the sequence 'settles down' as it progresses further.
Understanding sequence limits is crucial because even if a sequence doesn't necessarily reach a particular number, it can still "tend towards" or "approach" a value, which is significant in calculus and analysis for numerous applications.

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Most popular questions from this chapter

Laplace transforms. Let \(f\) b: continuous on \([0,\infty)\). The Laplace transform of \(f\) is the function \(F\) defined by setting $$F(s)=\int_{0}^{\infty} e^{-s x} f(x) d x$$ The domain of \(F\) is the set of numbers \(s\) for which the improper integral converges. Find the Laplace transform \(F\) of each of the following functions specifying the domain of \(F\). $$f(x)=e^{n x}$$

A ball is dropped from a height of 100 feet. Each time it hits the ground, it rebounds to \(75 \%\) of its previous height. (a) Let \(S_{n}\) be the distance that the ball travels between the \(n\) th and the \((n+1)\) st bounce. Find a formula for \(S_{n}\). (b) Let \(T_{n}\) be the time that the ball is in the air between the \(n\) th and the \((n+1)\) st bounce. Find a formula for \(T_{n}\).

Show that (a) \(\int_{-\infty}^{\infty} \sin x d x\) diverges although (b) \(\lim _{h \rightarrow \infty} \int_{-h}^{b} \sin x d x=0\).

Starting with \(0 < a < b,\) form the arithmetic mean \(a_{1}=\frac{1}{2}(a+b)\) and the geometric mean \(b_{1}=\sqrt{a b} .\) For \(n=2.3,4, \cdots\) set $$a_{n}=\frac{1}{2}\left(a_{n-1}+b_{n-1}\right) \quad and \quad b_{n}=\sqrt{a_{n-1} b_{n-1}}$$ (b) Show that the two sequences converge and \(\lim _{n \rightarrow \infty} a_{n}=\) \(\lim _{x \rightarrow \infty} b_{n} .\) The common value of this limit is called the aruhmetic-geontetric mean of a and b.

Use comparison test (11.7.2) to determine whether the integral converges. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$
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