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Chapter 9: Problem 14

Determine whether the given series must diverge because its terms do not converge to \(0 .\) $$ \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{n}\right) $$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{n}\right) \) does not diverge because the terms converge to 0.

Step by step solution

01

Understand the series

The given series is \( \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{n}\right) \). We must determine whether this series diverges by checking if its terms converge to 0.
02

Analyze the term \( a_n \)

The general term of the series is \( a_n = \left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{n}\right) \). As \( n \to \infty \), both \( \left(1 + \frac{1}{n}\right) \to 1\) and \( \ln \left(1 + \frac{1}{n}\right) \to 0\). We need to analyze what happens to the product of these two expressions.
03

Simplify the logarithmic term

For \( n \) large, we can approximate the logarithmic term using \( \ln (1 + x) \approx x \) when \( x \) is close to 0. For \( \ln \left(1 + \frac{1}{n}\right) \), this becomes approximately \( \frac{1}{n} \).
04

Evaluate the behavior of \( a_n \)

Substituting the approximation in, we get \( a_n \approx \left(1+\frac{1}{n}\right) \frac{1}{n} = \frac{1}{n} + \frac{1}{n^2} \). As \( n \to \infty \), this expression tends to \( \frac{1}{n} \).
05

Determine the limit of the term \( a_n \)

We have \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left( \frac{1}{n} + \frac{1}{n^2} \right) = \lim_{n \to \infty} \frac{1}{n} = 0 \). However, this only shows that \( a_n \to 0 \); it does not determine whether the series converges.

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Key Concepts

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

Convergence Testing
Convergence testing is crucial in analyzing whether an infinite series converges or diverges. In the context of our series, it involves examining if the terms of the series approach zero.
If the terms do not tend to zero, the series will definitely diverge.
  • We often employ tests like the Limit Comparison Test, the Ratio Test, or the Root Test to determine convergence.
  • These tests provide insights by establishing benchmarks or reference series to compare the original series against.
The goal is to determine the end behavior of the series鈥� terms and confirm whether they contribute meaningfully over an infinite number of summands. For our series, convergence testing starts with checking if the individual terms, denoted as \( a_n \), actually approach zero as \( n \to \infty \).
Sequence Limit
The sequence limit is a fundamental concept when analyzing series. It refers to the value that the terms of a sequence approach as the index \( n \) goes to infinity.
In the series provided, the term \( a_n = \left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{n}\right) \) is considered.
  • As \( n \to \infty \), the expression \( 1 + \frac{1}{n} \) approaches 1 and \( \ln \left(1 + \frac{1}{n}\right) \) approaches 0.
  • Utilizing logarithmic approximations and simplifying the expression to \( \frac{1}{n} + \frac{1}{n^2} \), we see the sequence approaches zero.
This is significant since the sequence limit influences the potential divergence or convergence of the entire series.
Infinite Series Analysis
Infinite series analysis explores whether the sum of an infinite list of terms results in a finite number or not. In our problem, the series is given by \( \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right) \ln \left(1+\frac{1}{n}\right) \).
Understanding the behavior of its terms, \( a_n \), is crucial.
  • If the terms do not tend to zero, the series diverges.
  • A series converging implies that these terms, when added up infinitely, approach a finite total.
In the current analysis, while \( a_n \to 0 \) as \( n \to \infty \), it still needs further testing beyond just this observation to conclude if the entire series converges.
Logarithmic Approximations
Logarithmic approximations simplify the analysis of series involving logarithms. For small values of \( x \), the logarithm \( \ln (1+x) \) can be approximated by \( x \). This simplifies mathematical manipulations.
In our given series, \( \ln \left(1 + \frac{1}{n}\right) \) becomes approximately \( \frac{1}{n} \) when \( n \) is large.
  • This approximation helps break down complex terms into easier-to-handle components.
  • While this approximation simplifies understanding term behavior, precise analysis requires more than basic approximations.
By applying these approximations, we gain clearer insight into whether the converging terms sufficiently approach zero, aiding in the broader task of determining series convergence or divergence.

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Most popular questions from this chapter

Let \(a \neq 0\), and assume that \(\lim _{n \rightarrow \infty} a_{n}=a\) and \(a_{n} \neq 0\) for all \(n\). Show that \(\sum_{n=1}^{\infty}\left|a_{n+1}-a_{n}\right|\) converges if and only if \(\sum_{n=1}^{\infty}\left|\frac{1}{a_{n+1}}-\frac{1}{a_{n}}\right|\) converges.

Suppose \(P\) dollars are deposited in a savings account at an interest rate of \(r\) percent per year, compounded \(n\) times a year. a. Show that the amount of money in the account after 1 year is given by $$R_{n}=P\left(1+\frac{0.01 r}{n}\right)^{n}$$ dollars (Hint: The amount after \(1 / n\) years is \(P(1+0.01 r / n) .)\) b. Find $$R=\lim _{n \rightarrow \infty} P\left(1+\frac{0.01 r}{n}\right)^{n}$$ This is the amount in the account after 1 year if the interest is compounded "continuously." c. If \(P=1000\) and \(r=5\), find the difference between the amounts after 1 year if the interest is compounded continuously and if it is compounded quarterly.

a. Show that $$ \begin{aligned} \frac{\tan ^{-1} t}{t} &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1} t^{2 n} \\\ &=1-\frac{t^{2}}{3}+\frac{t^{4}}{5}-\frac{t^{6}}{7}+\cdots \quad \text { for } 0<|t|<1 \end{aligned} $$ b. Using part (a), conclude that $$ \lim _{t \rightarrow 0}\left(\tan ^{-1} t\right) / t=1 $$

An indeterminately large number of identical blocks 1 unit long are stacked on top of each other. Show that it is possible for the top block to protrude as far from the bottom block as we wish without the blocks toppling (Figure 9.18). (Hint: The center of gravity of the top block must lie over the second block; the center of gravity of the top two blocks must lie over the third block, and so on. Thus the top block can protrude up to \(\frac{1}{2}\) unit from the end of the second block, the second block can protrude up to \(\frac{1}{4}\) unit from the end of the third block, the third block can protrude up to \(\frac{1}{6}\) unit from the end of the fourth block, and so on. Assuming that the center of gravity of the first \((n-1)\) blocks lies over the end of the \(n\) th block, show that the \(n\) th block can protrude up to \(1 / 2 n\) units from the end of the \((n+1)\) st block.)

If $$\$ 1000$$ is deposited in a savings account at an interest rate of \(r\) percent per year, then the number of dollars (principal plus interest) in the account after 1 year is \(1000(1+0.01 r) .\) Write a formula for the sequence that gives the amount of money in the account after \(n\) years for any positive integer \(n\).
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