/*! 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 14 Test the series for convergence ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 14

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty} \frac{\sin 2 n}{1+2^{n}}$$

Short Answer

Expert verified
The series \(\sum_{n=1}^{\infty} \frac{\sin 2n}{1+2^n}\) converges.

Step by step solution

01

Understand the Series

The given series is \(\sum_{n=1}^{\infty} \frac{\sin 2n}{1+2^n}\). We are tasked with determining whether this series converges or diverges. The terms of the series are \(a_n = \frac{\sin 2n}{1+2^n}\).
02

Check the General Term

Evaluate the behavior of the term \(a_n = \frac{\sin 2n}{1+2^n}\). Notice that \(|\sin 2n| \leq 1\), so \(|a_n| \leq \frac{1}{1+2^n}\). This suggests that the terms of the series are decreasing in magnitude and are bounded.
03

Apply the Limit Comparison Test

Compare the given series with the series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\), which is a geometric series with a common ratio \(r = \frac{1}{2}\) and is known to converge. Compute:\[\lim_{n \to \infty} \frac{\left| \frac{\sin 2n}{1+2^n} \right|}{\frac{1}{2^n}} = \lim_{n \to \infty} \frac{2^n \cdot |\sin 2n|}{1 + 2^n}\].
04

Evaluate the Limit

Simplify the expression:\[\lim_{n \to \infty} \frac{2^n \cdot |\sin 2n|}{1 + 2^n} = \lim_{n \to \infty} \frac{2^n}{2^n} = 1\].Since this limit is a positive finite number, and \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) converges, by the Limit Comparison Test, \(\sum_{n=1}^{\infty} \frac{\sin 2n}{1+2^n}\) also converges.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 a series. Imagine you have two series, \( \sum a_n \) and \( \sum b_n \), where you suspect that both of them can be compared. The main idea is to compute the limit:\[\lim_{n \to \infty} \frac{a_n}{b_n}\]
  • If this limit is a positive finite number (i.e., not zero or infinity), then both series behave the same in terms of convergence: both either converge or diverge.
  • The test is particularly useful when the original series is complex or not well understood, but it can be compared to a simpler series with known behavior.
In our example, the series \(\sum_{n=1}^{\infty} \frac{\sin 2n}{1+2^n}\) is compared to the series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\). By using the Limit Comparison Test, we found that the series converges since the limit of the ratio of their terms equals 1, a positive finite number.
Geometric Series
A geometric series is one of the simplest types of series you've likely encountered. It takes the form:\[\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + \ldots\]Where:
  • \(a\) is the first term of the series.
  • \(r\) is the common ratio that each term is multiplied by to get the next term.
Geometric series are special because their convergence is easy to determine. If the absolute value of the common ratio \(|r| < 1\), the geometric series converges. Otherwise, it diverges. In the original solution, the series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) is a converging geometric series with a common ratio \(r = \frac{1}{2}\). This fact was used effectively with the Limit Comparison Test to draw conclusions about the original series.
Bounded Sequences
A sequence \(a_n\) is said to be bounded if there exists a number \(M\) such that for all \(n\), the terms of the sequence are contained within the interval \( -M \leq a_n \leq M \). This essentially means that the sequence does not grow too large or drop too low. For convergence of series, having bounded terms can be helpful in conjunction with other tests:
  • In the given exercise, \(|\sin 2n| \leq 1\) ensures that the terms \( \frac{|\sin 2n|}{1 + 2^n} \) are bounded as well.
  • As a result, understanding bounded behavior was key in suggesting that the series might converge.
Combining knowledge of bounded behavior and efficient tests like the Limit Comparison Test offers a robust strategy for analyzing series convergence. In practical terms, noticing that the series鈥� terms were bounded provided an initial hint toward convergence before deeper analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Show that the function defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}}} & {\text { if } x \neq 0} \\\ {0} & {\text { if } x=0}\end{array}\right.$$ is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin.

If \(f(x)=\Sigma_{n-0}^{\infty} c_{a} x^{n},\) where \(c_{n+4}=c_{n}\) for all \(n \geqslant 0,\) find the interval of convergence of the series and a formula for \(f(x)\)

In Section 4.8 we considered Newton's method for approximating a root \(r\) of the equation \(f(x)=0,\) and from an initial approximation \(x_{1}\) we obtained successive approximations \(X_{2}, X_{3}, \ldots,\) where $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$$ Use Taylor's Inequality with \(n=1, a=x_{n},\) and \(x=r\) to show that if \(f^{\prime \prime}(x)\) exists on an interval I containing \(r, x_{n}\) and \(x_{n+1},\) and \(\left|f^{\prime \prime}(x)\right| \leqslant M,\left|f^{\prime}(x)\right| \geqslant K\) for all \(x \in I,\) then $$\left|x_{n+1}-r\right| \leqslant \frac{M}{2 K}\left|x_{n}-r\right|^{2}$$ [This means that if \(x_{n}\) is accurate to \(d\) decimal places, then \(x_{n+1}\) is accurate to about 2\(d\) decimal places. More precisely, if the error at stage \(n\) is at most \(10^{-m}\) , then the error at stage \(n+1\) is at most \((M / 2 K) 10^{-2 m} \cdot ]\)

Test the series for convergence or divergence. $$\sum_{j=1}^{\infty}(-1)^{j} \frac{\sqrt{j}}{j+5}$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=\sinh 2 x, \quad a=0, \quad n=5, \quad-1 \leqslant x \leqslant 1$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.