/*! 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 42 In the following exercises, use ... [FREE SOLUTION] | 91影视

91影视

In the following exercises, use an appropriate test to determine whether the series converges. $$ \sum_{n=1}^{\infty} \frac{(n+1)^{2}}{n^{3}+(1.1)^{n}} $$

Short Answer

Expert verified
The series converges by the Comparison Test.

Step by step solution

01

Choose a Test for Convergence

To determine if the series \( \sum_{n=1}^{\infty} \frac{(n+1)^2}{n^3 + (1.1)^n} \) converges, we will use the Comparison Test. This is because the series has a polynomial in the numerator and a combination of a polynomial and an exponential term in the denominator.
02

Simplifying the Expression

For large values of \( n \), notice that \( (1.1)^n \) grows significantly faster than \( n^3 \). Therefore, we can simplify the expression and focus on the dominant part of the denominator, which is \( (1.1)^n \). Hence, approximately \( \frac{(n+1)^2}{(1.1)^n} \).
03

Selecting a Comparable Series

Compare with the series \( \sum_{n=1}^{\infty} \frac{n^2}{(1.1)^n} \). Here, as \( n \rightarrow \infty \), \( (n+1)^2 \approx n^2 \), which makes this a good comparison.
04

Determine Convergence of Comparable Series

Inspect \( \sum_{n=1}^{\infty} \frac{n^2}{(1.1)^n} \). As \( 1.1^n \) grows exponentially faster than \( n^2 \), the individual terms \( \frac{n^2}{(1.1)^n} \rightarrow 0 \) rapidly. By employing the ratio test or analyzing the exponential decrements, we determine that this series converges.
05

Applying the Comparison Test

Since the comparison series \( \sum_{n=1}^{\infty} \frac{n^2}{(1.1)^n} \) converges and \( \frac{(n+1)^2}{n^3+(1.1)^n} \leq \frac{n^2}{(1.1)^n} \) for all sufficiently large \( n \), by the Comparison Test, the original series \( \sum_{n=1}^{\infty} \frac{(n+1)^2}{n^3 + (1.1)^n} \) 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.

comparison test
The Comparison Test is a handy tool when determining if a series converges. It works by comparing the series in question to another series with known convergence properties. If you have a new series you are uncertain about, you can use this test to compare it to a series whose behavior is already known.
Here's how it works: If the terms of your series are smaller than a known convergent series鈥 terms, your series must also converge. Conversely, if the terms are larger than those of a known divergent series, then your series diverges too.
To apply the Comparison Test effectively, choose a comparison series that is similar but simpler, ideally one for which the convergence or divergence is already established. This often involves focusing on dominant terms, especially ones that grow or shrink rapidly. The key is to simplify and approximate as needed.
polynomial and exponential terms
Polynomials and exponentials are two key components frequently seen in series. Understanding their roles is crucial when analyzing convergence.
In any function, polynomial terms grow at a slower rate than exponential terms as the variable goes to infinity. For example, if you have a series with a term like \( n^3 \) in the denominator, but also \( (1.1)^n \), the exponential part \( (1.1)^n \) will eventually dominate because it increases much faster.
When simplifying series, focus on the dominant term since it has the largest impact on the series' behavior. In many cases, exponentials can often simplify problems dramatically, by effectively overpowering polynomial parts for large values of \( n \). This means a term like \( n^3 + (1.1)^n \) will behave similarly to \( (1.1)^n \) when \( n \) is large.
convergence tests
Convergence tests are strategic tools for assessing whether series converge or diverge. There are several tests, each with its unique features, such as the Ratio Test, Root Test, Integral Test, and more. However, using the right test makes analysis easier and more efficient.
For series with exponential and polynomial functions, like our example, the Comparison Test is particularly useful. When chosen appropriately, these tests provide a straightforward approach to tricky infinite series.
Selecting the appropriate test depends largely on the given series' structure and behavior. It's key to identify dominant terms, simplify the expression, and choose a parallel series to compare with. Other tests might involve learning complex calculus concepts, but often a basic comparison does the trick.
exponential growth
Exponential growth refers to increases where the rate of growth is proportional to the current amount, leading to growth whose pace accelerates over time. This concept is important when dealing with series, as exponential terms, like \((1.1)^n\), become very large very quickly.
In the context of convergence, exponential growth in the denominator can be beneficial, as it causes terms to diminish rapidly, potentially leading the series to converge. Understanding this contrast between polynomial growth, which is slower and steadier, makes it easier to grasp why exponential terms dominate in determining the convergence behavior of a series.
  • Exponential terms dominate polynomial terms for large values of \( n \).
  • Even modest bases in exponential terms (like \(1.1\)) lead to significant growth.
  • Rapidly growing denominators effectively "shrink" term values, fostering convergence.

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

Use the root and limit comparison tests to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges. $$ \left.a_{n}=1 / x_{n}^{n} \text { where } x_{n+1}=\frac{1}{2} x_{n}+\frac{1}{x_{n}}, x_{1}=1 \text { (Hint: Find limit of }\left\\{x_{n}\right\\} .\right) $$

[T] To find an approximation for \(\pi\), set \(a_{0}=\sqrt{2+1}, a_{1}=\sqrt{2+a_{0}}\), and, in general, \(a_{n+1}=\sqrt{2+a_{n}}\). Finally, set \(p_{n}=3.2^{n} \sqrt{2-a_{n}}\). Find the first ten terms of \(p_{n}\) and compare the values to \(\pi\).

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{1}{4} \frac{3}{6} \frac{5}{8} \cdots \frac{2 n-1}{2 n+2}=\frac{1 \cdot 3-5 \cdots(2 n-1)}{2^{n}(n+1) !} .\) Explain why the ratio test cannot determine convergence of \(\sum_{n=1}^{\infty} a_{n} .\) Use the fact that \(1-1 /(4 k)\) is increasing \(k\) to estimate \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}\).

In the following exercises, use an appropriate test to determine whether the series converges. $$ \left.a_{k}=\left(\frac{k}{k+\ln k}\right)^{k} \text { (Hint: } a_{k}=\left(1+\frac{\ln k}{k}\right)^{-(k / \ln k) \ln k} \approx e^{-\ln k} \cdot\right) $$

For which values of \(r>0\), if any, does \(\sum_{n=1}^{\infty} r^{\sqrt{n}}\) converge? (Hint: \(\left.\sum_{n=1}^{\infty} a_{n}=\sum_{k=1}^{\infty} \sum_{n=k^{2}}^{(k+1)^{2}-1} a_{n} .\right)\)

See all solutions

Recommended explanations on Math Textbooks

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.