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Magazine Chapter 11: Problem 8
\(3-32\) Determine whether the series converges or diverges. $$\sum_{n=1}^{\infty} \frac{4+3^{n}}{2^{n}}$$
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Identify the Type of Series
The given series is \(\sum_{n=1}^{\infty} \frac{4+3^{n}}{2^{n}}\). This series has the form where the nth term is \(a_n = \frac{4+3^{n}}{2^{n}}\). We will analyze this expression to determine its convergence or divergence.
02
Simplify the General Term
Rewrite the term to separate the two components: \(a_n = \frac{4}{2^n} + \frac{3^n}{2^n}\). This can be further split into two series, \(\sum_{n=1}^{\infty} \frac{4}{2^n}\) and \(\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^n\), to be evaluated separately.
03
Analyze the First Series
The first series is \(\sum_{n=1}^{\infty} \frac{4}{2^n}\), which is a geometric series with the common ratio \(r = \frac{1}{2}\). Since \(|r| < 1\), this series converges. The sum of this series can be found as \(\frac{4}{1-\frac{1}{2}} = 8\).
04
Analyze the Second Series
The second series is \(\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^n\), which is also a geometric series, but with the common ratio \(r = \frac{3}{2}\). Since \(|r| > 1\), this series diverges.
05
Determine Overall Convergence
A series converges if all of its component series converge. In this case, \(\sum_{n=1}^{\infty} \frac{4}{2^n}\) converges, while \(\sum_{n=1}^{\infty} \left(\frac{3}{2}\right)^n\) diverges. Thus, the overall series \(\sum_{n=1}^{\infty} \left(\frac{4 + 3^n}{2^n}\right)\) diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a series of the form \( a + ar + ar^2 + \cdots \), where each term after the first is the product of the preceding term and a constant called the common ratio \( r \). The basic template for a geometric series involves:
- The first term \( a \)
- The common ratio \( r \)
Common Ratio
The common ratio \( r \) is the factor by which we multiply each term in a geometric series to get the next term. It's a critical component because it determines the behavior of the series:
- If the absolute value of \( r \) is less than one, \(|r| < 1\), the series will converge.
- If the absolute value of \( r \) is greater than or equal to one, \(|r| \geq 1\), the series will diverge.
Sum of Series
The sum of an infinite geometric series can be calculated using the formula \( \frac{a}{1-r} \). However, this formula only applies when the series converges, meaning \(|r| < 1\).In the given exercise, the first series \( \sum_{n=1}^{\infty} \frac{4}{2^n} \) had a common ratio \( r = \frac{1}{2} \), which allowed us to find its sum. We used \( a = 4 \) and calculated:\[\text{Sum} = \frac{4}{1 - \frac{1}{2}} = 8\]Understanding this formula is key to calculating the sum of convergent geometric series efficiently.
Infinite Series
An infinite series is a sum of infinitely many terms. Each term is a function of its position in the series, often following a specific pattern. These series can converge to a finite value or diverge to infinity.In our exercise, the expression \( \sum_{n=1}^{\infty} \left(\frac{4 + 3^n}{2^n}\right) \) represented an infinite series. Analyzing infinite series often requires us to separate them into simpler components. The series \( \frac{4}{2^n} \) converged due to its geometric nature and \( r < 1 \), while \( \left(\frac{3}{2}\right)^n \) diverged because \( r \geq 1 \). Understanding these distinctions helps us analyze and solve complex problems related to infinite sums.
