/*! 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 34 State whether each of the follow... [FREE SOLUTION] | 91影视

91影视

State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

Short Answer

Expert verified
The series converges conditionally.

Step by step solution

01

Rewrite the Series

Start by expressing the general term of the series: \((-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\). Observe that \((-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\) is an alternating series where each term follows the pattern of the difference of two fractions.
02

Simplify the General Term

The general term can be rewritten as \(a_n = \left(\frac{1}{n} - \frac{1}{n+1}\right) = \frac{1}{n(n+1)}\).This indicates that the series can be considered as:\(a_n = \frac{1}{n} - \frac{1}{n+1}\).Notice here that this part shows telescoping properties.
03

Check for Absolute Convergence

To check for absolute convergence, consider the absolute value of \(a_n\) without the alternating sign factor:\(\sum_{n=1}^{\infty} \left| \frac{1}{n(n+1)} \right|\).Since \(a_n = \frac{1}{n(n+1)}\) behaves like a telescoping series, we compare it with \(\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1} \right)\) which simplifies and is known to be convergent. Therefore, it doesn't absolutely converge.
04

Check for Conditional Convergence

Check if the original series converges using the alternating series test. Note:\((-1)^{n+1}\) introduces alternating behavior.For convergence of an alternating series:1. \(b_n = \frac{1}{n(n+1)}\) should decrease and approach 0 as \(n\rightarrow\infty\).Both conditions are satisfied making the series conditionally converging.

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.

Alternating Series
An alternating series is a series where the signs of its terms alternate between positive and negative. The series \[-(-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\]is an example of an alternating series because it includes the factor \((-1)^{n+1}\),which alternates the signs of the terms depending on whether \(n\) is odd or even. Alternating series can exhibit special properties that affect their convergence.

To determine if an alternating series converges, we use the Alternating Series Test. This test states that if the absolute values of the terms decrease steadily to zero, that is, \(b_{n+1} \leq b_n\)for all \(n\), and\(b_n \rightarrow 0\), the series converges.

In practical terms, alternating series often converge even if the series formed by the absolute values of the terms doesn't. Thus, identifying an alternating pattern in a series can be a helpful first step in determining its convergence properties.
Absolute Convergence
Absolute convergence refers to a series that converges even when all of its terms are replaced by their absolute values. For the series \[\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\],to check for absolute convergence, we consider the series formed by taking the absolute value of each term:\[\sum_{n=1}^{\infty} \left|\frac{1}{n} - \frac{1}{n+1}\right|.\]

In this case, the series becomes \[\sum_{n=1}^{\infty} \frac{1}{n(n+1)}.\]This series does not satisfy the requirements for absolute convergence because it behaves like a geometric series that doesn't converge when considering only its absolute values.

Understanding absolute convergence is crucial because it implies stability under changing terms. If a series converges absolutely, reordering its terms does not affect the sum. Unlike conditional convergence, this makes the series more stable and predictable.
Conditional Convergence
Conditional convergence occurs in series that converge only due to the alternating behavior of their terms, rather than their absolute values converging. In the case of \[\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right),\]we can apply the Alternating Series Test. The test verifies that the absolute values of terms decrease steadily and approach zero:
  • The sequence \(\frac{1}{n(n+1)}\) decreases as \(n\) increases.
  • As \(n\) approaches infinity, \(\frac{1}{n(n+1)} \to 0\).
Since these conditions are satisfied, the series converges conditionally.

Conditional convergence highlights the delicate balance that exists in certain series. Unlike absolute convergence, any reordering of terms in a conditionally convergent series can potentially change its sum. This property underscores the importance of understanding the series' structure before determining convergence.
Telescoping Series
A telescoping series derives its name from how its terms cancel out like a telescope collapsing. This series type often simplifies the process of finding its sum by reducing the expression within the series itself. The given series, with general term \[\frac{1}{n} - \frac{1}{n+1},\]exhibits this telescoping behavior.

When you write out the first few terms, you'll notice that most terms cancel out with subsequent ones:\[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots\]Only the initial term \(\frac{1}{1}\) survives because each term cancels with the next term in the series.

The magic of telescoping is its ability to transform an infinite series into a finite one for evaluation. Once telescoped, what's left is often straightforward to evaluate, making these series particularly interesting and useful in mathematical 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

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series \(\sum a_{k}\) with given terms \(a_{k}\) converges, or state if the test is inconclusive. $$ a_{k}=\frac{2-4 \cdot 6 \cdots 2 k}{(2 k) !} $$

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{\left(2^{n} n !\right)^{2}}{(2 n)^{2 n}} $$

[T] Find the first 1000 digits of \(\pi\) using either a computer program or Internet resource. Create a bit sequence \(b_{n}\) by letting \(b_{n}=1\) if the \(n\) th digit of \(\pi\) is odd and \(b_{n}=0\) if the \(n\) th digit of \(\pi\) is even. Compute the average value of \(b_{n}\) and the average value of \(d_{n}=\left|b_{n+1}-b_{n}\right|, n=1, \ldots, 999 .\) Does the sequence \(b_{n}\) appear random? Do the differences between successive elements of \(b_{n}\) appear random?

The following alternating series converge to given multiples of \(\pi .\) Find the value of \(N\) predicted by the remainder estimate such that the Nth partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum \(N\) for which the error bound holds, and give the desired approximate value in each case. Up to 15 decimals places, \(\pi=3.141592653589793 .\)[T] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form \(\sum_{n=1}^{\infty} \frac{S_{n}}{n}\), where \(s_{n}\) is a randomly generated sequence of \(\pm 1\) 's in which the values \(\pm 1\) are equally likely to occur. Use a random number generator to produce 1000 random \(\pm 1\) s and plot the partial sums \(S_{N}=\sum_{n=1}^{N} \frac{s_{n}}{n}\) of your random harmonic sequence for \(N=1\) to \(1000 .\) Compare to a plot of the first 1000 partial sums of the harmonic series.

Is the series convergent or divergent? $$ \sum_{n=1}^{\infty} n^{-(n+1 / n)} $$

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.