/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Which of the sequences \(\left\\... [FREE SOLUTION] | 91Ó°ÊÓ

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Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. \(a_{n}=\ln n-\ln (n+1)\)

Short Answer

Expert verified
The sequence \(a_n = \ln n - \ln (n+1)\) converges, and its limit is 0.

Step by step solution

01

Simplify the Sequence

The sequence given is \(a_n = \ln n - \ln (n+1)\). We can simplify this expression using the properties of logarithms. Specifically, we can apply the property \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\). Applying this property, we have \(a_n = \ln \left( \frac{n}{n+1} \right)\).
02

Find the Limit

Next, we need to find the limit of the sequence as \(n\) approaches infinity. \(a_n = \ln \left( \frac{n}{n+1} \right)\). As \(n\) approaches infinity, \(\frac{n}{n+1} = \frac{n}{n(1 + \frac{1}{n})} = \frac{1}{1 + \frac{1}{n}}\). As \(n\) becomes very large, \(\frac{1}{n}\) approaches 0, leading \(\frac{n}{n+1}\) to approach 1.
03

Evaluate the Limit Result

Since \(\frac{n}{n+1} \to 1\) as \(n \to \infty\), we have \(a_n = \ln \left( \frac{n}{n+1} \right) \to \ln(1)\). Evaluating the limit, we find that \(\ln(1) = 0\). Thus, the sequence \(a_n\) converges to 0.

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Key Concepts

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

Limit of a Sequence
When analyzing sequences, one important concept to understand is the **limit of a sequence**. The limit of a sequence is a value that the elements of the sequence approach as the index goes to infinity. In simpler terms, it is where the sequence "settles down" or becomes stable.
If such a value exists, the sequence is said to converge to that limit. Otherwise, the sequence diverges. In the exercise, the sequence is defined as \(a_n = \ln \left( \frac{n}{n+1} \right)\).
To determine convergence, we analyzed what happens to the expression \(\frac{n}{n+1}\) as \(n\) grows larger. Simplifying it, we found that \(\frac{n}{n+1}\) approaches \(1\) when \(n\) approaches infinity. Consequently, the sequence converges to the limit \(\ln(1)\), which is \(0\).
Properties of Logarithms
Logarithms are powerful mathematical tools that help us manipulate exponential expressions. One useful property is the difference of logs: \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\).
This property allows us to simplify expressions involving logarithms, as seen in the exercise. Originally, \(a_n = \ln n - \ln (n+1)\).
Using the log property, this expression became \(\ln \left( \frac{n}{n+1} \right)\). This reduced form is much easier to analyze when finding convergence.
  • It saves time by simplifying calculations.
  • Makes the problem more tractable.
  • Allows for easier limits evaluation.
Mastering these properties can greatly benefit students in solving a wide range of mathematical problems.
Infinite Limits
Infinite limits relate to the behavior of a sequence or function as it approaches infinity. It’s crucial in understanding whether sequences or functions grow without bound or stabilize at a specific value.
In the given problem, we needed to find the behavior of \(\frac{n}{n+1}\) as \(n\) becomes large.
Rewriting this expression as \(\frac{1}{1+\frac{1}{n}}\) reveals its behavior as \(n\to\infty\):
  • \(\frac{1}{n} \to 0\), thus \(1 + \frac{1}{n} \to 1\).
  • The entire fraction \(\frac{n}{n+1} \to 1\).
The logarithm of \(1\) is \(0\), indicating the sequence converges. Understanding infinite limits is essential because it often reveals hidden patterns and outcomes of sequences and functions as they stretch infinitely.

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Most popular questions from this chapter

a. Series for \(\sinh ^{-1} x \quad\) Find the first four nonzero terms of the Taylor series for $$ \sinh ^{-1} x=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}} $$ b. Use the first three terms of the series in part (a) to estimate \(\sinh ^{-1} 0.25 .\) Give an upper bound for the magnitude of the estimation error.

Assume that each sequence converges and find its limit. \(a_{1}=0, \quad a_{n+1}=\sqrt{8+2 a_{n}}\)

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2\). Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

Two complex numbers \(a+i b\) and \(c+\) id are equal if and only if \(a=c\) and \(b=d .\) Use this fact to evaluate $$ \int e^{a x} \cos b x d x \quad \text { and } \quad \int e^{a x} \sin b x d x $$ from $$ \int e^{(a+i b) x} d x=\frac{a-i b}{a^{2}+b^{2}} e^{(a+i b) x}+C $$ where \(C=C_{1}+i C_{2}\) is a complex constant of integration.

The approximation \(e^{x}=1+x+\left(x^{2} / 2\right)\) is used when \(x\) is small. Use the Remainder Estimation Theorem to estimate the error when \(|x|<0.1\)

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