/*! 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 35 Which of the series in Exercises... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the series in Exercises \(11-44\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} $$

Short Answer

Expert verified
The series converges absolutely because the series of absolute values converges.

Step by step solution

01

Identify the series

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \). This is an alternating series due to the presence of \((-1)^n\).
02

Test for absolute convergence

To test for absolute convergence, consider the series \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| = \sum_{n=1}^{\infty} \frac{(n+1)^n}{(2n)^n} \). Simplify the terms: \( \frac{(n+1)^n}{(2n)^n} = \left(\frac{n+1}{2n}\right)^n = \left(\frac{1}{2} + \frac{1}{2n}\right)^n \). As \(n \to \infty\), this expression tends to \( (\frac{1}{2})^n \). This is a geometric series with ratio \(r = \frac{1}{2} < 1\), hence it converges.
03

Conclusion about absolute convergence

Since the series of absolute values \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| \) converges, the original series converges absolutely.

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 91Ó°ÊÓ!

Key Concepts

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

Alternating Series
An alternating series is one where the terms alternate in sign. This type of series often takes the form\[ \sum (-1)^n a_n \]where the terms switch between positive and negative. The presence of \((-1)^n\) in the series results in this alternating sign. When analyzing such a series, one useful tool is the Alternating Series Test.
To apply this test, two conditions must be satisfied:
  • The absolute value of the terms, \(|a_n|\), must be decreasing.
  • The limit of \(a_n\) as \(n\) approaches infinity must be zero: \(\lim_{n \to \infty} a_n = 0\).
If both conditions are satisfied, the alternating series converges. However, convergence does not necessarily imply absolute convergence, which leads us to our next core concept.
Absolute Convergence
A series is said to converge absolutely if the series of its absolute values converges. In other words, a series \( \sum a_n \) converges absolutely if \( \sum |a_n| \) converges.
For example, if we consider the alternating series given in the exercise, the absolute value series is: \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| = \sum_{n=1}^{\infty} \frac{(n+1)^n}{(2n)^n} \]This transformation removes alternating signs, simplifying our analysis.
Absolute convergence is important because it implies ordinary convergence. If a series converges absolutely, it will also converge. However, a series can converge without converging absolutely.
Geometric Series
A geometric series is a series of the form\[ \sum_{n=0}^{\infty} ar^n \]where \(a\) is the first term and \(r\) is the common ratio. The series converges if \(|r| < 1\) and diverges if \(|r| \geq 1\).
In our analysis of the series' absolute convergence, we found that the absolute series closely resembles a geometric series:\[ \left( \frac{1}{2} + \frac{1}{2n} \right)^n \]which approaches \((\frac{1}{2})^n\) as \(n\) goes to infinity. Here, the common ratio \(r\) is \(\frac{1}{2}\), which is less than 1, indicating convergence.
Recognizing a series as geometric makes it simple to determine its behavior, using the criterion \(|r| < 1\) for convergence.
Ratio Test
The ratio test is a powerful tool for determining convergence or divergence of an infinite series. It involves examining the limit of the absolute ratio of successive terms\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
If the limit \(L < 1\), the series converges absolutely. If \(L > 1\), or is infinite, the series diverges. If \(L = 1\), the test is inconclusive, and another method must be used.
While the explicit use of the ratio test wasn't detailed in the exercise, recognizing when it can apply, such as when dealing with terms of the form \(b_n^n\) where \(b_n\) varies with \(n\), is crucial. This test can simplify finding convergence properties for many series types, complementing the insights gained from geometric or alternating series tests.

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. Series for sinh \(^{-1} x\) 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.

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) $$ \sum_{n=3}^{\infty} \frac{(1 / n)}{(\ln n) \sqrt{\ln ^{2} n-1}} $$

In Exercises \(121-124\) , experiment with a calculator to find a value of \(N\) that will make the inequality hold for all \(n>N\) . Assuming that the inequality is the one from the formal definition of the limit of a sequence, what sequence is being considered in each case and what is its limit? $$ 2^{n} / n !<10^{-7} $$

Series for sin \(^{-1} x\) Integrate the binomial series for \(\left(1-x^{2}\right)^{-1 / 2}\) to show that for \(|x|<1\) , $$\quad \sin ^{-1} x=x+\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot 6 \cdots \cdot(2 n)} \frac{x^{2 n+1}}{2 n+1}$$

The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\}\).

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.