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Chapter 9: Problem 63

Show that \(\Sigma_{n=2}^{\infty}\left((\ln n)^{q} / n^{p}\right)\) converges for \(-\infty1\) (Hint: Limit Comparison with \(\Sigma_{n=2}^{\infty} 1 / n^{\prime}\) for \(1

Short Answer

Expert verified
The series converges for any real \( q \) and \( p > 1 \).

Step by step solution

01

Rewrite the Series

Consider the series \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \). We want to determine if this series converges. The hint suggests using the limit comparison test with the series \( \sum_{n=2}^{\infty} \frac{1}{n^{r}} \) where \(1 < r < p\).
02

Choose a Comparison Series

Select the comparison series \( \sum_{n=2}^{\infty} \frac{1}{n^{r}} \), which is convergent for \( r > 1 \). Here, \( r \) is chosen such that \( 1 < r < p \). This series will help us apply the limit comparison test.
03

Apply the Limit Comparison Test

Compute the limit \( L = \lim_{n \to \infty} \frac{\left( \frac{(\ln n)^{q}}{n^{p}} \right)}{\left( \frac{1}{n^{r}} \right)} = \lim_{n \to \infty} \frac{(\ln n)^{q} n^{r}}{n^{p}} = \lim_{n \to \infty} (\ln n)^{q} n^{r-p} \).
04

Determine the Behavior of the Limit

Since \( p > r \), the term \( n^{r-p} \) behaves like \( \frac{1}{n^{p-r}} \), which tends to zero as \( n \to \infty \). Additionally, as \( n \to \infty \), \( (\ln n)^{q} \) grows slower than any polynomial power of \( n \). Hence, \( L = 0 \).
05

Conclusion from the Limit Comparison

Since the limit \( L = 0 \), which is a finite number and \( \sum_{n=2}^{\infty} \frac{1}{n^{r}} \) converges, we conclude by the limit comparison test that \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \) also converges.

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Key Concepts

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

Convergence of Series
Understanding the convergence of series is crucial in mathematical analysis. A series is said to converge if the sum of its infinite terms approaches a finite number as more terms are added.
For a series like \( \sum_{n=2}^{\infty} \frac{(\ln n)^q}{n^p} \), convergence depends on the growth rate of the numerator \((\ln n)^q\) compared to the denominator \(n^p\).
  • If the denominator grows faster than the numerator, the terms become smaller, leading to convergence.
  • If the numerator grows faster or at the same rate, the series may diverge.
Tests like the Limit Comparison Test help determine the convergence by comparing it to a known convergent series. This approach standardizes proof techniques and provides a methodological solution to analytic problems.
Logarithmic Functions
Logarithmic functions, denoted as \( \ln n \) for natural log, are important in evaluating the growth of functions. They exhibit very slow growth compared to polynomial functions.
  • The logarithmic function increases indefinitely as \( n \to \infty \), but at a much lesser rate than any positive power of \( n \).
  • This characteristic is what makes \((\ln n)^q\) less dominant in series when combined with polynomial terms in the denominator.
When analyzing series incorporating log functions, it is important to note this property, especially in convergence tests where the polynomial part may overshadow the logarithmic component.
Power Series
Power series are infinite sums in which each term is a power of \( n \), usually expressed in the form \( \sum_{n=0}^{\infty} a_n x^n \). They play a crucial role in calculus and real analysis.
In our case, we are not dealing directly with a power series but the concepts intersect, especially when considering terms like \( n^p \).
  • The power \( p \) determines how rapidly the terms of a series shrink, influencing convergence.
  • A larger exponent in the denominator compared to the numerator generally signals convergence.
Understanding how different powers influence series growth helps in analyzing the limits and behavior of series in analytic proofs.
Mathematical Proofs
Mathematical proofs are logical arguments that demonstrate the truth of a statement or theorem. In the context of series convergence, proofs are necessary to justify that a series has a finite sum.
The Limit Comparison Test is a powerful tool used in these proofs. It involves:
  • Selecting a series with known behavior which serves as a benchmark.
  • Calculating the limit of the ratio of the terms of the given series to the terms of the comparison series.
  • Using the outcome of this limit to conclude if the original series converges, diverges, or requires further investigation.
These steps create a framework that addresses how one series can reflect the properties of another, a foundational principle in advanced calculus and analysis.

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Most popular questions from this chapter

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{4^{n} 2^{n} n !}$$

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{3^{n}}{n^{3}}$$

Assume that \(b_{n}\) is a sequence of positive numbers converging to 1/3. Determine whether the following series converge or diverge. a. \(\sum_{n=1}^{\infty} \frac{b_{n+1} b_{n}}{n 4^{n}}\) b. \(\sum_{n=1}^{\infty} \frac{n^{n}}{n ! b_{1}^{2} b_{2}^{2} \cdots b_{n}^{2}}\)

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=2}^{\infty} \frac{n}{(\ln n)^{(n / 2)}}$$

Which of the series, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{[2 \cdot 4 \cdot \cdots \cdot(2 n)]\left(3^{n}+1\right)}$$
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