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Using the divergence test For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive. a. \(\sum_{n=1}^{\infty} \frac{n}{3 n-1}\) b. \(\sum_{n=1}^{\infty} \frac{1}{n^{3}}\) C. \(\sum_{n=1}^{\infty} e^{1 / n^{2}}\)

Short Answer

Expert verified
(a) Diverges, (b) Inconclusive, (c) Diverges.

Step by step solution

01

Understand the Divergence Test

The divergence test states that if the limit of the sequence \(a_n\) in the series \(\sum a_n\) is not zero, then the series diverges. Specifically, if \(\lim_{n \to \infty} a_n eq 0\), then the series diverges. If the limit is zero, the divergence test is inconclusive, and the series may still converge or diverge.
02

Apply the Divergence Test to Series (a)

First, identify the sequence \(a_n = \frac{n}{3n-1}\). To apply the divergence test, compute the limit:\[\lim_{n \to \infty} \frac{n}{3n - 1} = \lim_{n \to \infty} \frac{1}{3 - \frac{1}{n}} = \frac{1}{3}\]Since \(\frac{1}{3} eq 0\), the series diverges according to the divergence test.
03

Apply the Divergence Test to Series (b)

Identify the sequence \(a_n = \frac{1}{n^3}\). Compute the limit:\[\lim_{n \to \infty} \frac{1}{n^3} = 0\]Since the limit equals zero, the divergence test is inconclusive for this series. Further tests are needed to determine convergence.
04

Apply the Divergence Test to Series (c)

Identify the sequence \(a_n = e^{1/n^2}\). To apply the divergence test, compute the limit:\[\lim_{n \to \infty} e^{1/n^2} = e^0 = 1\]Since \(1 eq 0\), the series diverges according to the divergence test.

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Key Concepts

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

Infinite Series
Infinite series are sequences of numbers that continue indefinitely. These are written as the sum of their terms, often denoted by the sigma symbol: \(\sum_{n=1}^{\infty} a_n\).Each term in the series is part of a sequence, and you explore this by working on the limit as the number of terms approaches infinity.
An important part of analyzing an infinite series is determining whether it converges or diverges.Most often, students use this to understand complex functions or calculate approximations.It's crucial to know different tests and methods to evaluate these series.
Here's a quick rundown:
  • An infinite series converges if the sum approaches a finite limit.
  • If the terms do not approach zero, or if the series does not approach a fixed sum, it diverges.
Divergence does not mean the series has a particular pattern. It just means it does not settle to a finite value.Different tests like the divergence test, ratio test, and others help to ascertain whether a series converges or diverges.
Convergence
Convergence in infinite series is a critical concept in calculus, as it determines the behavior of the series towards a limit.A convergent series means its partial sums approach a specific number as more terms of the series are added.
Mathematically, a series \(\sum_{n=1}^{\infty} a_n\) converges if for some number \(L\), you can make the partial sum \(S_n\) as close to \(L\) as desired by taking more terms from the sequence.To determine convergence, individuals must use various tests, each designed for different series.
Some common tests are:
  • Divergence Test: Checks if the limit of the sequence of terms is not zero. If it is not zero, the series diverges.
  • Ratio Test: Often used for series with factorials or exponential functions.
  • Root Test: Useful for series where each term is raised to a power.
Knowing which test to apply comes from the series’ structure. The key is ensuring effective convergence analysis through practice and familiarity.
Sequence Limit
The sequence limit is foundational for understanding infinite series and their convergence.This concept involves finding what value a sequence \(a_n\) approaches as \(n\) grows indefinitely.If \(\lim_{n \to \infty} a_n = L\), the sequence converges to \(L\).
It's important to analyze this limit because it serves as a preliminary check for divergence in series.When examining a series like \(\sum a_n\), the first step is confirming whether \(\lim_{n \to \infty} a_n = 0\).This check is part of the divergence test, allowing students to quickly ascertain if a series can potentially converge.
The sequence limit is easy when dealing with simple sequences, but complex series require algebraic manipulation or l'Hôpital's rule to evaluate limits.This proves advantageous in solving practical math problems in calculus and other mathematical disciplines.Determine convergence is simplified by understanding the concept of limits deeply, allowing you to apply it across varied mathematical challenges.

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Most popular questions from this chapter

The following series converge by the ratio test. Use summation by parts, \(\sum_{k=1}^{n} a_{k}\left(b_{k+1}-b_{k}\right)=\left[a_{n+1} b_{n+1}-a_{1} b_{1}\right]-\sum_{k=1}^{n} b_{k+1}\left(a_{k+1}-a_{k}\right)\), to find the sum of the given series. $$ \left.\sum_{k=1}^{\infty} \frac{k}{c^{k}}, \text { where } c>1 \text { (Hint: Take } a_{k}=k \text { and } b_{k}=c^{1-k} /(c-1) .\right) $$

[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 Euler transform rewrites \(S=\sum_{n=0}^{\infty}(-1)^{n} b_{n}\) as \(S=\sum_{n=0}^{\infty}(-1)^{n} 2^{-n-1} \sum_{m=0}^{n}\left(\begin{array}{c}n \\ m\end{array}\right) b_{n-m}\). For the alternating harmonic series, it takes the form \(\ln (2)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=\sum_{n=1}^{\infty} \frac{1}{n 2^{n}} .\) Compute partial sums of \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\) until they approximate \(\ln (2)\) accurate to within \(0.0001\). How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate \(\ln (2)\).

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.If \(b_{n} \geq 0\) and \(\lim _{n \rightarrow \infty} b_{n}=0\) then \(\sum_{n=1}^{\infty}\left(\frac{1}{2}\left(b_{3 n-2}+b_{3 n-1}\right)-b_{3 n}\right)\) converges.

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

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