Problem 32
Question
In Exercises \(27-34,\) use the \(n\) th-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. $$ \sum_{n=0}^{\infty} \frac{e^{n}}{e^{n}+n} $$
Step-by-Step Solution
Verified Answer
The series diverges by the n-th term test as \( \lim_{{n \to \infty}} a_n = 1 \neq 0 \).
1Step 1: Identify the n-th term of the series
The given series is \( \sum_{n=0}^{\infty} \frac{e^{n}}{e^{n}+n} \). The n-th term of the series, denoted by \( a_n \), is \( a_n = \frac{e^{n}}{e^{n}+n} \). Our task is to determine the limit of \( a_n \) as \( n \to \infty \).
2Step 2: Calculate the limit of the n-th term
Compute \( \lim_{{n \to \infty}} a_n \). Begin by dividing the numerator and the denominator of \( a_n \) by \( e^n \):\[a_n = \frac{e^n}{e^n + n} = \frac{1}{1 + \frac{n}{e^n}}.\] As \( n \to \infty \), the term \( \frac{n}{e^n} \to 0 \) because the exponential function \( e^n \) grows much faster than the linear term \( n \). Therefore,\[\lim_{{n \to \infty}} a_n = \frac{1}{1+0} = 1.\]
3Step 3: Apply the n-th term test for divergence
The n-th term test for divergence states that if \( \lim_{{n \to \infty}} a_n eq 0 \), then the series \( \sum a_n \) is divergent. Since \( \lim_{{n \to \infty}} a_n = 1 eq 0 \), the series \( \sum_{n=0}^{\infty} \frac{e^{n}}{e^{n}+n} \) diverges by the n-th term test.
Key Concepts
Divergent SeriesExponential FunctionsLimits and Convergence
Divergent Series
A series is termed as divergent if it does not settle down to a specific value as more and more terms are added. Instead, it continues to grow indefinitely. The n-th Term Test for Divergence is one way to determine whether a series is divergent. Here's how it works:
We consider the sequence \(a_n\) that makes up the series. The series \(\sum a_n\) is divergent if the limit of its n-th term, \(\lim_{{n \to \infty}} a_n\), is not equal to zero.
This works because, intuitively, if the terms of a series don't approach zero, their sum cannot stabilize to a particular value. In our provided series example, the n-th term is \(a_n = \frac{e^n}{e^n + n}\). As calculated, this limit is equal to \(1\), showing us the series is divergent. A convergent series, in contrast, would have terms that diminish to zero as more terms are added, potentially approaching a fixed sum.
We consider the sequence \(a_n\) that makes up the series. The series \(\sum a_n\) is divergent if the limit of its n-th term, \(\lim_{{n \to \infty}} a_n\), is not equal to zero.
This works because, intuitively, if the terms of a series don't approach zero, their sum cannot stabilize to a particular value. In our provided series example, the n-th term is \(a_n = \frac{e^n}{e^n + n}\). As calculated, this limit is equal to \(1\), showing us the series is divergent. A convergent series, in contrast, would have terms that diminish to zero as more terms are added, potentially approaching a fixed sum.
Exponential Functions
Exponential functions are in the form \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to \(2.71828\). Exponential functions are known for their unique property of growing rapidly. This rapid growth makes them very useful in various fields, including compounding interest, population growth, and radioactive decay.
One of the most significant characteristics of exponential functions is their rate of increase. For example, when facing a series like \(e^n/(e^n + n)\), the exponential within the fraction grows significantly faster than the linear term \(n\), allowing us to evaluate limits effectively. This behavior becomes evident when analyzing the provided series, where \(e^n\) quickly overpowers \(n\) as \(n\) becomes large, making the fraction approach \(1\) as mentioned earlier.
Understanding exponential functions helps with various mathematical applications, including solving complex growth models and analyzing financial data.
One of the most significant characteristics of exponential functions is their rate of increase. For example, when facing a series like \(e^n/(e^n + n)\), the exponential within the fraction grows significantly faster than the linear term \(n\), allowing us to evaluate limits effectively. This behavior becomes evident when analyzing the provided series, where \(e^n\) quickly overpowers \(n\) as \(n\) becomes large, making the fraction approach \(1\) as mentioned earlier.
Understanding exponential functions helps with various mathematical applications, including solving complex growth models and analyzing financial data.
Limits and Convergence
A limit is a tool that describes the behavior of a function or sequence as it approaches a particular point. Evaluating limits helps us to predict the behavior of the function and to determine the convergence or divergence of a series.
Convergence refers to a series or sequence settling to a fixed value. For example, if a sequence approaches a particular number as its terms tend toward infinity, it converges. Conversely, if it does not approach any number, it diverges.
In the context of our example series, the n-th term was calculated to find that \(\lim_{{n \to \infty}} a_n = 1\). Since the limit is not zero, the series diverges. Understanding convergence and divergence helps to solve complex problems and predict long-term behavior in mathematical models.
Convergence refers to a series or sequence settling to a fixed value. For example, if a sequence approaches a particular number as its terms tend toward infinity, it converges. Conversely, if it does not approach any number, it diverges.
In the context of our example series, the n-th term was calculated to find that \(\lim_{{n \to \infty}} a_n = 1\). Since the limit is not zero, the series diverges. Understanding convergence and divergence helps to solve complex problems and predict long-term behavior in mathematical models.
- Convergent series: Series where \(\lim_{n \to \infty} a_n = 0\).
- Divergent series: Series where \(\lim_{n \to \infty} a_n eq 0\).
Other exercises in this chapter
Problem 32
Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\
View solution Problem 32
In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) condition
View solution Problem 33
Use series to evaluate the limits. \begin{equation} \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} \end{equation}
View solution Problem 33
Find the first four nonzero terms in the Maclaurin series for the functions in Exercises \(29-34 .\) $$e^{\sin x}$$
View solution