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Chapter 11: Problem 62

Graph, on the same coordinate axes, \(y=x\) and \(y=\) \((\ln x)^{k}\) for \(k=1,2,3\) and \(1 \leq x \leq 200\), and then use the graphs to predict whether the series \(\sum_{n=1}^{x} \frac{1}{(\ln n)^{k}}\) converges or diverges for \(k=1,2,\) and \(3 .\)

Short Answer

Expert verified
The series diverges for \( k = 1 \) and converges for \( k = 2, 3 \).

Step by step solution

01

Understand and Plot the Functions

We need to graph the functions: \( y = x \) and \( y = (\ln x)^k \) for \( k=1, 2, 3 \) over the range \( 1 \leq x \leq 200 \). First, understand that \( y = x \) is a linear function through the origin. The function \( y = (\ln x)^k \) changes behavior based on the value of \( k \). When \( x \) is small, \( \ln x \) is negative for \( x < 1 \) and stays below \( y = x \), but when \( x \) becomes large, the natural logarithm increase logarithmically which affects the compete between \( y=x \) and \( y=(\ln x)^k \) depends on whether \( k \).
02

Analyze the Graphs for Comparison

Plot \( y = (\ln x)^1 = \ln x \), \( y = (\ln x)^2 \), and \( y = (\ln x)^3 \) alongside \( y = x \). When \( 1 \leq x \leq 200 \), \( y = \ln x \) will grow slower than \( y = x \), and this gap increases as \( k \) increases. For \( y = (\ln x)^2 \) and \( y = (\ln x)^3 \), they initially grow slowly compared to \( y = x \), but the effect becomes more significant as \( x \) increases, especially for larger \( k \). This discrepancy helps in understanding whether the series diverges or converges.
03

Use Graphs to Assess Series Convergence/Divergence

Consider the series \( \sum_{n=1}^{\infty} \frac{1}{(\ln n)^k} \). Compare it to the harmonic series \( \sum \frac{1}{n} \), which is known to diverge. If \( (\ln x)^k \) grows significantly slower than \( x \) as \( x \to \infty \), then the term \( \frac{1}{(\ln x)^k} \) remains relatively large and similar to the harmonic term, suggesting potential divergence. For \( k = 1 \), the series resembles a harmonic series, hence diverges. For \( k = 2, 3, \ldots \), \( (\ln x)^k \) grows faster than \( log x \), leading \( \sum \frac{1}{(\ln x)^2} \) and higher powers to converge.
04

Conclude Based on Observations

From the graph and comparison with known series, we infer: for \( k = 1 \), the series diverges as \( \frac{1}{\ln n} \) behaves similarly to \( \frac{1}{n} \). For \( k > 1 \), \((\ln n)^k\) grows faster causing \( \frac{1}{(\ln n)^k} \) to decrease faster, thus converging similarly to \( \sum \frac{1}{n^2} \). Hence for \( k = 2, 3 \), the series converges.

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Key Concepts

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

Natural Logarithm Function
The natural logarithm function, denoted as \( \ln x \), is a fundamental concept in mathematics, especially in calculus and analysis. It's the inverse function of the exponential function \( e^x \). The natural logarithm of a number \( x \) answers the question: "To what power must \( e \) be raised to yield \( x \)?". Understanding its properties helps when trying to compare and analyze functions such as \( y = (\ln x)^k \).
  • Behavior for Small \( x \): For values of \( x \) close to 1, \( \ln x \) is quite small. As \( x \) approaches 1 from the right, \( \ln x \) approaches 0.
  • Growth at Larger \( x \): The function grows slowly and steadily as \( x \) increases beyond 1, because the increase follows a logarithmic scale.
  • Curve Nature: The graph of \( \ln x \) starts off with a steep increase but gradually becomes less steep as \( x \) increases.
In the context of series, \( (\ln x)^k \) behaves differently depending on \( k \), impacting convergence and divergence analysis for terms like \( \frac{1}{(\ln n)^k} \).
Harmonic Series
The harmonic series is a pivotal concept when discussing convergence and divergence of infinite series. It is defined as \( \sum \frac{1}{n} \). This series is known to diverge, meaning that as you keep summing its terms, the total can grow arbitrarily large.
  • Comparison With \( y = (\ln x)^k \): For \( k = 1 \), \( \sum \frac{1}{(\ln n)^k} \) resembles the harmonic series and also diverges since \( \ln x \) grows slower than \( x \).
  • Convergence for Larger \( k \): When \( k > 1 \), \( (\ln x)^k \) grows faster than \( \ln x \), causing the series \( \sum \frac{1}{(\ln n)^k} \) to converge, similar to \( \sum \frac{1}{n^2} \).
Understanding how harmonic series diverges helps in determining whether other similar series may potentially converge or diverge depending on their growth rate behavior.
Graphical Analysis of Functions
Graphical analysis involves plotting functions to visually assess their behavior over certain intervals. In the exercise, comparing functions \( y = x \) and \( y = (\ln x)^k \) through graphs helps in understanding their growth relative to each other.
  • Linear vs. Logarithmic Growth: While \( y = x \) is a straightforward linear increase, \( y = \ln x \) grows logarithmically, much slower in comparison.
  • Effect of Exponent \( k \): Increasing \( k \) in \( (\ln x)^k \) affects its graph significantly, making \( y = (\ln x)^k \) increase faster compared to \( \ln x \), especially for large \( x \).
  • Visualization: By graphing, you can see the intersection and relative gap between these curves, which indicates how the term \( \frac{1}{(\ln x)^k} \) will behave.
Visualizing these functions on the same axes allows for a clearer understanding of how their growth rates affect the convergence or divergence of corresponding series.

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Most popular questions from this chapter

Light is absorbed by rods and cones in the retina of the eye. The number of photons absorbed by a photoreceptor during a given flash of light is governed by the Poisson distribution. More precisely, the probability \(p_{n}\) that a photoreceptor absorbs exactly \(n\) photons is given by the formula \(p_{n}=e^{-\lambda} \lambda^{n} / n !\) for some \(\lambda>0\). |al Show that \(\sum_{n=0}^{n} p_{n}=1\). |b| Sight usually occurs when two or more photons are absorbed by a photoreceptor. Show that the probability that this will occur is \(1-e^{-\lambda}(\lambda+1)\)

Find the first three terms of the Taylor series for \(f(x)\) at \(c\). \(f(x)=\tan ^{-1} x ; \quad c=1\)

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Use the first two nonzero terms of a Maclaurin series to approximate the number, and estimate the error in the approximation. \(\tan ^{-1} 0.1\)

Find the radius of convergence of the power series. $$ \sum_{n=0}^{x} \frac{(n+1) !}{10^{n}}(x-5)^{n} $$
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