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Chapter 11: Problem 33

Which of the series in Exercises 1鈥�36 converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}} $$

Short Answer

Expert verified
The series converges since it resembles the convergent p-series \( \sum \frac{1}{n^2} \).

Step by step solution

01

Write the General Term

The general term of the series is given as \( a_n = \frac{1}{n \sqrt[n]{n}} \), where \( n \) is a positive integer starting from 1.
02

Simplify the General Term

Recognize that \( \sqrt[n]{n} = n^{1/n} \). Hence, the term becomes \( a_n = \frac{1}{n^{1 + \frac{1}{n}}} \).
03

Apply the Limit Comparison Test

For large \( n \), the value of \( n^{1/n} \) approaches 1. Therefore, \( a_n = \frac{1}{n \cdot n^{1/n}} \approx \frac{1}{n^2} \) for large \( n \). Compare this with the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which converges because \( p = 2 > 1 \).
04

Conclusion Using Limit Comparison Test

Because the general term \( a_n \) of our series behaves like \( \frac{1}{n^2} \) for large \( n \), and since the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, the given series also converges.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a powerful tool in determining the convergence or divergence of an infinite series. It helps when you suspect the series can be compared to a simpler series, often a p-series. Here鈥檚 how it works:
  • Take the general term of your series, say \( a_n \), and pick a comparator series term, say \( b_n \), that shares similar characteristics.
  • Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
  • If this limit is a finite positive number, then both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
In our example, the term \( a_n = \frac{1}{n^{1 + \frac{1}{n}}} \) was simplified and compared with the term of the p-series \( \frac{1}{n^2} \). Since the limit comparison showed a finite positive constant, both series converge.
p-Series
A p-series is a classic form of series written as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant. The behavior of this series solely relies on the value of \( p \):
  • If \( p > 1 \), the series converges.
  • If \( p \le 1 \), the series diverges.
The p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which has \( p = 2 \), is one of the most well-known convergent series. Its convergence is often used as a benchmark for comparison, as seen in the Limit Comparison Test used for our specific series.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. Think of it as adding up numbers without end. The primary interest in infinite series is determining whether the sum reaches a finite number (converges) or keeps growing without limit (diverges). This concept is fundamental in advanced mathematics and calculus.

To work with infinite series, certain tests, such as the comparison tests, ratio test, and root test are frequently used. These tests help evaluate if an infinite process of adding results in a meaningful total.
General Term Simplification
Simplifying the general term of an infinite series is often a necessary step before applying any convergence tests. Simplification allows for easier comparison with known series types like p-series.
For example, in \( a_n = \frac{1}{n \sqrt[n]{n}} \), recognizing that \( \sqrt[n]{n} = n^{1/n} \) helps restate the term as \( a_n = \frac{1}{n^{1 + \frac{1}{n}}} \). This new form is closer to the typical p-series which aids in analysis.
Simplification not only makes the term more navigable but also paves the way for identifying similarities with other series, making it a crucial step in solving such problems.

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Most popular questions from this chapter

Outline of the proof of the Rearrangement Theorem (Theo- rem 17\()\) a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n},\) and let \(s_{k}=\sum_{n=1}^{k} a_{n}\) . Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\) $$\sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \quad\( and \)\quad\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2}$$ since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\},\) there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3},\) then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1} .\) Therefore, if \(n \geq N_{3}\) $$\begin{aligned}\left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned}$$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\)

a. Show that $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3} ?\) Give reasons for your answer.

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2} .\)

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=x, \quad y^{\prime}(0)=1 \text { and } y(0)=2 $$
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