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Chapter 8: Problem 37

\(37-40=\) Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? $$a_{n}=\frac{1}{2 n+3}$$

Short Answer

Expert verified
The sequence is decreasing and bounded between 0 and \(\frac{1}{3}\).

Step by step solution

01

Define the Sequence Behavior

Given the sequence \(a_n = \frac{1}{2n + 3}\). To determine if the sequence is increasing, decreasing, or not monotonic, we need to analyze the behavior of \(a_n\) as \(n\) increases. Observe that as \(n\) increases, the denominator \(2n + 3\) also increases, causing the value of \(a_n\) to decrease because the numerator is constant.
02

Confirm Decreasing Behavior Mathematically

To mathematically confirm that the sequence is decreasing, consider the terms \(a_n\) and \(a_{n+1}\). We have:\[ a_n = \frac{1}{2n + 3} \]\[ a_{n+1} = \frac{1}{2(n+1) + 3} = \frac{1}{2n + 5} \]Since \(2n + 3 < 2n + 5\), it follows that \(\frac{1}{2n+3} > \frac{1}{2n+5}\), thus \(a_n > a_{n+1}\). Therefore, the sequence is indeed decreasing.
03

Determine Boundedness

For boundedness, observe that as \(n\) tends to infinity, \(2n + 3\) tends to infinity, which causes \(a_n = \frac{1}{2n + 3}\) to tend to zero. Thus, the sequence is bounded below by zero. Also, since \(a_n\) is a positive fraction, the sequence is bounded above by \(\frac{1}{3}\), which is the value of \(a_0\). Therefore, the sequence is bounded between 0 and \(\frac{1}{3}\).

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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 said to be monotonic if it consistently increases or decreases. Understanding the behavior of a sequence is key to determining its monotonicity. In the case of our sequence, \(a_n = \frac{1}{2n + 3}\), we observe how each term changes as \(n\) increases. Since the denominator \(2n + 3\) gets larger, the fraction itself becomes smaller, indicating a decrease in value. Thus, our sequence is decreasing.

Monotonic sequences can be classified into two types:
  • Increasing Sequences: where each term is greater than the one before it.
  • Decreasing Sequences: where each term is less than the one before it, as is the case for \(a_n\).
If a sequence alternates directions or has no clear pattern, it's termed non-monotonic.
Bounded Sequences
A sequence is bounded if its values lie within a certain range and do not go to infinity or negative infinity. For our sequence \(a_n = \frac{1}{2n + 3}\), we considered what happens as \(n\) approaches infinity. The denominator increases, causing \(a_n\) to approach zero, the sequence's lower bound.

Being bounded implies two key traits:
  • Lower Bound: The smallest value that a sequence can approach. In our case, \(a_n\) cannot decrease below zero.
  • Upper Bound: The largest value the sequence can attain. For \(a_n\), it is \(\frac{1}{3}\), the value when \(n = 0\).
By these observations, our sequence lies between 0 and \(\frac{1}{3}\), confirming it is bounded.
Decreasing Sequences
Decreasing sequences are a specific type of monotonic sequence where each term is smaller than the one before it. This behavior is clearly seen in \(a_n = \frac{1}{2n + 3}\). Here, by comparing successive terms, namely \(a_n\) and \(a_{n+1}\), we note that \(\frac{1}{2n + 3} > \frac{1}{2n + 5}\). This indicates that each term in the sequence is less than its predecessor.

To determine if a sequence is decreasing, we:
  • Compare consecutive terms to ensure \(a_n > a_{n+1}\).
  • Show mathematically that this pattern holds for all terms within the sequence.
This process verifies that the sequence's value declines over time, characterizing it as a decreasing sequence.

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Most popular questions from this chapter

(a) Suppose that \(\Sigma a_{n}\) and \(\Sigma b_{n}\) are series with positive terms and \(\Sigma b_{n}\) is convergent. Prove that if \(\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0\) then \(\Sigma a_{n}\) is also convergent. (b) Use part (a) to show that the series converges. $$\sum_{n=1}^{\infty} \frac{\ln n}{n^{3}} \quad \text { (ii) } \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} $$

If the radius of convergence of the power series \(\Sigma_{n-0}^{\infty} c_{n} x^{n}\) is \(10,\) what is the radius of convergence of the series \(\sum_{n=1}^{\infty} n c_{n} x^{n-1} ?\) Why?

Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int x^{2} \ln (1+x) d x $$

Express the number as a ratio of integers. \(10.1 \overline{35}=10.135353535 \ldots\)

Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{1}{x+10} $$
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