Problem 13
Question
Verify that the infinite series diverges. $$ \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} $$
Step-by-Step Solution
Verified Answer
The given infinite series \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \) simplifies to \( \sum_{n=1}^{\infty} \big(\frac{1}{2} + \frac{1}{2^{n+1}}\big)\). The limit of this sequence as \(n\) approaches infinity is not zero, therefore according to the divergence test, the series diverges.
1Step 1: Identify the series
The series provided is \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \). This is an infinite series because it extends to infinity.
2Step 2: Simplify the series
The series can be rewritten by combining like terms in the numerator. Divide both terms in the numerator by \(2^{n+1}\), giving the simplified series \( \sum_{n=1}^{\infty} \big(\frac{2^{n}}{2^{n+1}} + \frac{1}{2^{n+1}}\big) \). This simplifies further to \( \sum_{n=1}^{\infty} \big(\frac{1}{2} + \frac{1}{2^{n+1}}\big)\). This series clearly separates into two series: one is a constant and the other is a geometric series with a ratio of \( 1/2 \).
3Step 3: Evaluate the limit of the sequence
Consider the limit as \(n\) approaches infinity for the sequence \( a_n = \frac{1}{2} + \frac{1}{2^{n+1}} \). The second term vanishes as \( n \) approaches infinity, hence the limit will be \( \frac{1}{2} \), which is not equal to zero.
4Step 4: Apply the divergence test
Since the limit of the sequence of terms of the series is not zero, by the divergence test, the infinite series is divergent.
Key Concepts
Divergence TestGeometric SeriesLimit of a SequenceInfinite Series Simplification
Divergence Test
The Divergence Test is a fundamental tool in identifying whether an infinite series diverges. For an infinite series \( \sum a_n \), if the limit of the sequence \( a_n \) as \( n \) approaches infinity is not zero, then the series diverges. This test is straightforward: it simply checks the individual terms of the sequence that make up the series.
If \( \lim_{{n \to \infty}} a_n eq 0 \), the terms do not approach zero, and the series fails to converge to a finite sum. In this case, as demonstrated for the series \( \sum_{n=1}^{\infty} \left( \frac{1}{2} + \frac{1}{2^{n+1}} \right) \), the limit was \( \frac{1}{2} \), which clearly isn’t zero. Hence, the series diverges. Always remember, if the Divergence Test proves that \( \lim_{{n \to \infty}} a_n = 0 \), it does not automatically imply convergence but tells us the series might converge or require further testing.
If \( \lim_{{n \to \infty}} a_n eq 0 \), the terms do not approach zero, and the series fails to converge to a finite sum. In this case, as demonstrated for the series \( \sum_{n=1}^{\infty} \left( \frac{1}{2} + \frac{1}{2^{n+1}} \right) \), the limit was \( \frac{1}{2} \), which clearly isn’t zero. Hence, the series diverges. Always remember, if the Divergence Test proves that \( \lim_{{n \to \infty}} a_n = 0 \), it does not automatically imply convergence but tells us the series might converge or require further testing.
Geometric Series
A geometric series is a type of series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The general form is \( a + ar + ar^2 + ar^3 + \cdots \), or more conventionally \( \sum_{n=0}^{\infty} ar^n \).
In the exercise, part of the series is identified as a geometric series: \( \sum_{n=1}^{\infty} \frac{1}{2^{n+1}} \). Here, the common ratio \( r \) is \( \frac{1}{2} \), which is between \(-1\) and \(1\), indicating convergence of this part. However, this insight helps in understanding the structure but does not alter the fact that the original series diverges due to the constant \( \frac{1}{2} \). Geometric series can provide insights into part of an infinite series' behavior, leading to a deeper understanding of its convergence properties.
In the exercise, part of the series is identified as a geometric series: \( \sum_{n=1}^{\infty} \frac{1}{2^{n+1}} \). Here, the common ratio \( r \) is \( \frac{1}{2} \), which is between \(-1\) and \(1\), indicating convergence of this part. However, this insight helps in understanding the structure but does not alter the fact that the original series diverges due to the constant \( \frac{1}{2} \). Geometric series can provide insights into part of an infinite series' behavior, leading to a deeper understanding of its convergence properties.
Limit of a Sequence
In dealing with infinite series, understanding the limit of a sequence is crucial. A sequence is an ordered list of numbers, and as \( n \) goes to infinity, the sequences in question approach a fixed value, or limit. For the series \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \), we consider the sequence
\( a_n = \frac{1}{2} + \frac{1}{2^{n+1}} \).
As \( n \rightarrow \infty \), \( \frac{1}{2^{n+1}} \) approaches zero because the denominator grows exponentially, hence the sequence approaches \( \frac{1}{2} \). Knowing the limit of a sequence helps in determining whether the terms of the series diminish appropriately for possible convergence.
\( a_n = \frac{1}{2} + \frac{1}{2^{n+1}} \).
As \( n \rightarrow \infty \), \( \frac{1}{2^{n+1}} \) approaches zero because the denominator grows exponentially, hence the sequence approaches \( \frac{1}{2} \). Knowing the limit of a sequence helps in determining whether the terms of the series diminish appropriately for possible convergence.
Infinite Series Simplification
Infinite series simplification involves rewriting series in a form that makes evaluating convergence or divergence easier. This involves algebraic manipulation to separate terms or factor expressions as needed.
In the problem, the initial series \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \) was simplified to \( \sum_{n=1}^{\infty} \left( \frac{1}{2} + \frac{1}{2^{n+1}} \right) \). This step of simplification helps identify components within the series—such as constant terms and geometric sequences—that are treated separately to understand their contribution to the series' overall behavior. Simplifying series can reveal interrelationships between parts of the expression, providing a clearer path to convergence analysis.
In the problem, the initial series \( \sum_{n=1}^{\infty} \frac{2^{n}+1}{2^{n+1}} \) was simplified to \( \sum_{n=1}^{\infty} \left( \frac{1}{2} + \frac{1}{2^{n+1}} \right) \). This step of simplification helps identify components within the series—such as constant terms and geometric sequences—that are treated separately to understand their contribution to the series' overall behavior. Simplifying series can reveal interrelationships between parts of the expression, providing a clearer path to convergence analysis.
Other exercises in this chapter
Problem 12
Determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} $$
View solution Problem 12
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ g(x)=\frac{4 x-7}{2 x^{2}+3 x-2}, \quad c=0 $$
View solution Problem 13
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{2 n}{n^{2}+1} $$
View solution Problem 13
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=x e^{x}, \quad n=4 $$
View solution