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Chapter 8: Problem 56

Determine whether the following series converge or diverge. $$\sum_{k=1}^{\infty} \frac{2^{k}+3^{k}}{4^{k}}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Write out the Ratio Test Formula

The ratio test is used to determine the convergence or divergence of a series by calculating the limit of the ratio of consecutive terms in the series. The formula for the ratio test is: $$\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = L$$ Where \(a_k\) is the k-th term of the series and \(L\) is the limit of the ratio of consecutive terms.
02

Apply the Ratio Test Formula to the Given Series

The given series is: $$\sum_{k=1}^{\infty} \frac{2^{k}+3^{k}}{4^{k}}$$ So we have \(a_k = \frac{2^k + 3^k}{4^k}\). We want to find the limit of the ratio of consecutive terms, so we need to calculate \(a_{k+1}\) and then find the limit of \(\frac{a_{k+1}}{a_k}\) as \(k \to \infty\): $$a_{k+1} = \frac{2^{k+1} + 3^{k+1}}{4^{k+1}}$$
03

Calculate the Limit of the Ratio

Now we need to find the limit of \(\frac{a_{k+1}}{a_k}\) as \(k \to \infty\): $$\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim_{k \to \infty} \frac{\frac{2^{k+1} + 3^{k+1}}{4^{k+1}}}{\frac{2^k + 3^k}{4^k}} = \lim_{k \to \infty} \frac{2^{k+1} + 3^{k+1}}{4^{k+1}} \cdot \frac{4^k}{2^k + 3^k}$$ Simplify the expression: $$\lim_{k \to \infty} \frac{(2^{k+1} + 3^{k+1})(4^k)}{(4^{k+1})(2^k + 3^k)}$$ Divide both numerator and denominator by \(4^k\): $$\lim_{k \to \infty} \frac{(2^{k+1} + 3^{k+1})}{(4^{k+1})(2^k + 3^k)} \cdot \frac{1}{4^k} = \lim_{k \to \infty} \frac{(2^{k+1} + 3^{k+1})}{4(2^{k+1} + 3^{k+1})}$$ Simplify further: $$\lim_{k \to \infty} \frac{1}{4}$$
04

Determine the Convergence or Divergence of the Series

The limit of the ratio \(\frac{a_{k+1}}{a_k}\) is \(\frac{1}{4}\). According to the ratio test: - If \(L < 1\), the series converges absolutely. - If \(L > 1\), the series diverges. - If \(L = 1\), the test is inconclusive. Since \(L = \frac{1}{4} < 1\), the series converges absolutely. Therefore, the given series $$\sum_{k=1}^{\infty} \frac{2^{k}+3^{k}}{4^{k}}$$ converges.

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Key Concepts

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

Ratio Test
The Ratio Test is a valuable tool for determining whether an infinite series converges or diverges. Its power lies in analyzing the behavior of consecutive terms in a series. The test focuses on the limit of the ratio formed by dividing one term by the previous term in a series. More specifically, it uses the formula:
  • \( \lim_{k \to \infty} \frac{a_{k+1}}{a_k} = L \)
Where \( a_k \) is the k-th term, and \( L \) is the resulting limit after applying the formula. To apply the Ratio Test, the calculated \( L \) must be compared against the number 1 with the following rules:
  • If \( L < 1 \), the series converges absolutely, meaning it will sum to a definite value.
  • If \( L > 1 \), the series diverges, lacking a finite sum.
  • If \( L = 1 \), the test is inconclusive, requiring use of other convergence criteria.
Using this straightforward test, determining the nature of a series becomes a process of calculation and comparison.
Infinite Series
In mathematics, an infinite series is a sum of an infinite sequence of terms. Unlike finite series, infinite series do not end; they continue indefinitely by definition. This ongoing nature often leads to questions about their sum and behavior.The series is typically denoted as \( \sum_{k=1}^{\infty} a_k \), where each \( a_k \) represents a term in the sequence. The significance of an infinite series hinges on whether it converges or diverges. Convergence occurs when the series approaches a specific value as more terms are added. Divergence indicates that no finite value is reached.Various tests and methods, including the Ratio Test, serve as tools to analyze infinite series' behavior. Their thorough examination resides at the heart of calculus and mathematical analysis, assisting in unraveling the mysteries of endless sums.
Convergence Tests
Convergence tests are essential techniques used to determine the behavior of an infinite series. They help us understand whether a series converges鈥攖hat is, approaches a finite sum鈥攐r diverges, where no such sum exists. Here are some commonly used tests:- **Ratio Test**: Evaluates the ratio of sequential terms and concludes based on the resulting limit, as previously discussed.- **Root Test**: Similar to the Ratio Test, but uses \( \sqrt[k]{|a_k|} \) to gauge behavior as \( k \) increases.Besides these, other tests exist, each suitable for various types of series. These include the Integral Test, Comparison Test, and Alternating Series Test, to name a few. The choice of test depends on the nature of the series at hand and the specific problem needing a solution. Such tests allow mathematicians and students alike to navigate through intricate series with greater confidence and clarity.

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Most popular questions from this chapter

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty}(-1)^{k}$$

\(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \ldots\right\\} .\) Assume that \(a_{k}>0\) for all \(k\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of \(P=\lim _{n \rightarrow \infty} P_{n}\) c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.
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