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Chapter 9: Problem 11

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=1}^{\infty} \frac{n}{n+200} $$

Short Answer

Expert verified
The series diverges by the Divergence Test.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{n}{n+200} \). It is an infinite series with terms \( a_n = \frac{n}{n+200} \). This series resembles a rational series.
02

Simplify the General Term

Simplify the term \( a_n = \frac{n}{n+200} \) by dividing the numerator and the denominator by \( n \): \[ a_n = \frac{1}{1 + \frac{200}{n}}. \] As \( n \to \infty \), \( \frac{200}{n} \to 0 \), thus \( a_n \to 1 \).
03

Apply the Divergence Test

The Divergence Test states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum a_n \) diverges. Here, \( \lim_{n \to \infty} \frac{1}{1 + \frac{200}{n}} = 1 \). Since this limit is not zero, according to the Divergence Test, the series diverges.

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Key Concepts

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

Divergence Test
The Divergence Test is a simple yet powerful tool for checking the behavior of an infinite series. In essence, this test helps determine whether a series diverges by examining the limit of its general term.

It states that if the limit of the sequence of terms of a series, denoted as \(a_n\), does not equal zero \(\lim_{n \to \infty} a_n eq 0\), then the series \(\sum a_n\) must diverge.

Always remember:
  • If \(\lim_{n \to \infty} a_n = 0\), the series has a chance to converge, but it doesn鈥檛 guarantee convergence.
  • If \(\lim_{n \to \infty} a_n eq 0\), the series definitely diverges.
In our example, when discussing the series \(\sum_{n=1}^{\infty} \frac{n}{n+200}\), applying the Divergence Test was straightforward: the limit of the terms approaches 1. Since 1 is not zero, the series diverges.
Infinite Series
An infinite series represents the sum of infinitely many terms, written in the form \(\sum_{n=1}^{\infty} a_n\). It is essentially the limit of the partial sums of a given sequence as the number of terms approaches infinity.

Key points about infinite series:
  • The sequence \(a_n\) is called the sequence of terms.
  • The partial sum \(S_n\) is the sum of the first \(n\) terms \(S_n = a_1 + a_2 + \ldots + a_n\).
  • The infinite series converges if the partial sums approach a fixed number as \(n\) approaches infinity.
  • Conversely, a series diverges if the partial sums do not approach a fixed number.
In mathematical problems, like the one we are discussing, the distinction between convergence and divergence of an infinite series is crucial, as it determines the behavior of the series over an unbounded range of terms.
Rational Series
Rational series involve terms that can be expressed as a ratio of two polynomial functions. In the given exercise, we see a typical rational series where the term is \(\frac{n}{n+200}\).

This type of series can often be simplified for easier analysis, primarily through algebraic manipulation.

Here's why rational series are significant:
  • They frequently appear in calculus and are often used in test problems for convergence.
  • Understanding how to manipulate these series is key to applying various convergence tests, like the Divergence Test.
In our case, simplifying \(\frac{n}{n+200}\) by dividing both the numerator and denominator by \(n\) gives \(\frac{1}{1 + \frac{200}{n}}\). As \(n\) increases, \(\frac{200}{n}\) shrinks towards zero, making the behavior of the series more apparent. This simplification is crucial for evaluating the limit and determining divergence or convergence.
Limit of a Sequence
The limit of a sequence is a foundational concept in understanding the behavior of series. It refers to the value that the terms of a sequence \(a_n\) approach as \(n\) becomes very large.

Essentially, it describes the "end behavior" of the sequence.
When evaluating the limit of a sequence:
  • Consider what happens to \(a_n\) as \(n\to\infty\).
  • If the terms approach a specific number, the limit is that number.
  • If the terms do not stabilize to a single value, the limit does not exist.
In the provided series having \(a_n = \frac{1}{1 + \frac{200}{n}}\), evaluating the limit as \(n\) approaches infinity shows us an important detail: this simplifies to 1, which informs us that the series does not converge to zero. Thus, it helps us, in conjunction with the Divergence Test, to determine the divergence of the series.

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Most popular questions from this chapter

Find the Taylor polynomial of order 3 based at a for the given function. $$ \sqrt{x} ; a=2 $$

One can sometimes find a Maclaurin series by the method of equating coefficients. For example, let $$ \tan x=\frac{\sin x}{\cos x}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots $$ Then multiply by \(\cos x\) and replace \(\sin x\) and \(\cos x\) by their series to obtain $$ \begin{aligned} x-\frac{x^{3}}{6}+\cdots &=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots\right)\left(1-\frac{x^{2}}{2}+\cdots\right) \\ &=a_{0}+a_{1} x+\left(a_{2}-\frac{a_{0}}{2}\right) x^{2}+\left(a_{3}-\frac{a_{1}}{2}\right) x^{3}+\cdots \end{aligned} $$ Thus, $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}-\frac{a_{0}}{2}=0, \quad a_{3}-\frac{a_{1}}{2}=-\frac{1}{6}, \quad \cdots $$ so $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}=0, \quad a_{3}=\frac{1}{3}, \quad \cdots $$ and therefore $$ \tan x=0+x+0+\frac{1}{3} x^{3}+\cdots $$ which agrees with Problem 1 . Use this method to find the terms through \(x^{4}\) in the series for \(\sec x\).

Plot on the same axes the given function along with the Maclaurin polynomials of orders \(1,2,3\), and \(4 .\) $$ e^{-x^{2}} $$

If an object of rest mass \(m_{0}\) has velocity \(v\), then (according to the theory of relativity) its mass \(m\) is given by \(m=\) \(m_{0} / \sqrt{1-v^{2} / c^{2}}\), where \(c\) is the velocity of light. Explain how physicists get the approximation $$ m \approx m_{0}+\frac{m_{0}}{2}\left(\frac{v}{c}\right)^{2} $$

Find the Taylor series in \(x-a\) through the term \((x-a)^{3}\). \(e^{x}, a=1\)
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