Chapter 9: Problem 18
Use differentiation to show that the given sequence is strictly increasing or strictly decreasing. $$ \left\\{\frac{\ln (n+2)}{n+2}\right\\}_{n=1}^{+\infty} $$
Short Answer
Expert verified
The sequence is strictly decreasing for \( n \geq 1 \).
Step by step solution
01
Define the Related Function
Consider the function associated with the sequence: \( f(x) = \frac{\ln(x+2)}{x+2} \). This function corresponds to the sequence values evaluated at integer points \( n \). To determine if the sequence is increasing or decreasing, we will analyze the derivative \( f'(x) \).
02
Find the Derivative
Differentiate \( f(x) = \frac{\ln(x+2)}{x+2} \) using the quotient rule. The quotient rule states that if \( h(x) = \frac{u(x)}{v(x)} \), then \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(x) = \ln(x+2) \) and \( v(x) = x+2 \).
03
Apply the Quotient Rule
Calculate the derivatives: \( u'(x) = \frac{1}{x+2} \) and \( v'(x) = 1 \). Substitute into the quotient rule formula: \[f'(x) = \frac{(\frac{1}{x+2})(x+2) - (\ln(x+2))(1)}{(x+2)^2} = \frac{1 - \ln(x+2)}{(x+2)^2}. \]
04
Analyze the Sign of Derivative
Determine where \( f'(x) > 0 \) or \( f'(x) < 0 \). The derivative \( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2} \) has the positive numerator \( 1 - \ln(x+2) > 0 \) when \( \ln(x+2) < 1 \). This implies \( x+2 < e \), thus \( x < e - 2 \).
05
Identify Critical Points
Since we are interested in integer values starting from \( n=1 \), consider \( n+2 < e \). With \( e \approx 2.718 \), \( n+2 < 2.718 \) leads to \( n < 0.718 \). This condition isn't met for natural numbers, hence \( 1 - \ln(n+2) < 0 \) for \( n \geq 1 \).
06
Conclude Increasing or Decreasing Behavior
Since \( f'(n) = \frac{1 - \ln(n+2)}{(n+2)^2} < 0 \) for all natural numbers \( n \geq 1 \), the sequence \( \left\{ \frac{\ln(n+2)}{n+2} \right\}_{n=1}^{\infty} \) is strictly decreasing.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Strictly Increasing Sequences
When a sequence is called strictly increasing, each term in the sequence is greater than the one before it. Think of it like climbing a staircase where every step you take is higher than the last. Mathematically, if you have a sequence \( a_n \), it is strictly increasing if \( a_{n+1} > a_n \) for all \( n \).
To determine if a sequence is strictly increasing, you check whether the related function’s derivative is positive over the interval you are interested in. If the derivative \( f'(x) > 0 \) for all \( x \) in the interval, it means the function, and therefore the sequence at integer points, climbs upwards continuously.
To determine if a sequence is strictly increasing, you check whether the related function’s derivative is positive over the interval you are interested in. If the derivative \( f'(x) > 0 \) for all \( x \) in the interval, it means the function, and therefore the sequence at integer points, climbs upwards continuously.
- Checking the sign of the derivative can help you decide if you're going up or down.
- Positive derivative means increasing sequence.
Strictly Decreasing Sequences
A sequence is strictly decreasing when every term is less than the one before it. It's like descending a staircase where each step down brings you lower. In mathematical terms, a sequence \( a_n \) is strictly decreasing if \( a_{n+1} < a_n \) for all \( n \).
The method to prove a strictly decreasing sequence involves checking the sign of the derivative of the associated function. When the derivative \( f'(x) < 0 \), the function descends, and so does the sequence.
The method to prove a strictly decreasing sequence involves checking the sign of the derivative of the associated function. When the derivative \( f'(x) < 0 \), the function descends, and so does the sequence.
- Negative derivative indicates the sequence is strictly decreasing.
- This method simplifies the problem by translating it into the language of calculus through differentiation.
Quotient Rule
The quotient rule is a key concept in calculus used to differentiate functions that are ratios of two differentiable functions. The rule is expressed as: if \( h(x) = \frac{u(x)}{v(x)} \), then \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).
This might seem complex at first glance, but breaking it down:
This might seem complex at first glance, but breaking it down:
- Identify \( u(x) \) and \( v(x) \), the functions in the numerator and denominator.
- Find their derivatives, \( u'(x) \) and \( v'(x) \).
- Plug these into the quotient rule formula to find \( h'(x) \).