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Converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$\sum_{n=1}^{\infty} \frac{2^{n}}{3^{n}}$$

Short Answer

Expert verified
The series converges as it is a geometric series with a common ratio less than 1.

Step by step solution

01

Recognize the form of the series

We need to determine whether the series \( \sum_{n=1}^{\infty} \frac{2^n}{3^n} \) converges or diverges. Recognize that this series takes the form of a geometric series. A geometric series is of the form \( \sum_{n=1}^{\infty} ar^{n} \) where \( a \) is the first term and \( r \) is the common ratio.
02

Find the first term and common ratio

For the series \( \sum_{n=1}^{\infty} \frac{2^n}{3^n} \), rewrite it as \( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n \). Here, the first term \( a \) is \( \frac{2}{3} \), and the common ratio \( r \) is also \( \frac{2}{3} \).
03

Determine convergence using geometric series test

A geometric series \( \sum_{n=1}^{\infty} ar^{n} \) converges if the absolute value of the common ratio \( |r| < 1 \). In our series, \( r = \frac{2}{3} \), and since \( |\frac{2}{3}| < 1 \), the series converges.
04

State the conclusion

Since the common ratio \( \frac{2}{3} \) satisfies the condition \( |r| < 1 \), the series \( \sum_{n=1}^{\infty} \frac{2^n}{3^n} \) is a convergent geometric series.

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Key Concepts

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

Series Convergence
When we talk about series convergence, we are referring to whether the series adds up to a finite number. This is especially important for infinite series, which continue indefinitely. If the sum of an infinite series results in a finite number, the series is said to converge. If not, it diverges. With geometric series, such as the one given in our problem, convergence is determined by evaluating the absolute value of the common ratio, denoted as \( r \).
  • If \(|r| < 1\), the series converges.
  • If \(|r| \geq 1\), the series diverges.
For the series \( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n \), we calculated \( r = \frac{2}{3} \). Since \( | \frac{2}{3} | \) is less than 1, the series converges. Convergence indicates that there is a specific value that the series approaches as the number of terms increases.
Common Ratio
The common ratio is a key component in geometric series. It is the factor by which we multiply each term to get the next term in the sequence. In mathematical notation for a geometric series, the common ratio is represented as \( r \).

For the series \( \sum_{n=1}^{\infty} \frac{2^n}{3^n} \), it can be rewritten as \( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n \). Here, the common ratio \( r \) is \( \frac{2}{3} \).
  • A common ratio \( |r| < 1 \) means the terms get smaller and smaller, allowing the series to converge.
  • A common ratio \( |r| \geq 1 \) means the terms do not decrease quickly enough, so the series might diverge.
Understanding the common ratio helps in quickly determining the behavior of the geometric series.
Infinite Series
Infinite series can appear daunting because they imply an endless process of adding terms together. However, they are a fundamental part of calculus and mathematical analysis. An infinite series is expressed as \( \sum_{n=1}^{\infty} a_n \), and it can either converge to a finite sum or diverge.

In geometric series, such as \( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n \), we focus on the common ratio to decide the series' fate—whether it converges or diverges. The unique property of an infinite series lies in its ability to sum an infinite number of terms and still result in a finite value if convergent.

Infinite series have many applications, including in complex calculations where they help in approximating functions and solving mathematical problems that involve sums of infinite terms. In daily life, they pop up in various scientific and engineering contexts.
  • A convergent infinite series remains manageable and predictable.
  • A divergent one does not, offering unlimited growth or oscillation.
Understanding their convergence properties is crucial to harnessing their power effectively.

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Most popular questions from this chapter

Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: a. For what values of \(x\) can the function be replaced by each approximation with an error less than \(10^{-2} ?\). b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step \(I:\) Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(x=0\) Step 3: Calculate the \((n+1)\) st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M .\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

If \(\Sigma a_{n}\) converges and \(a_{n}>0\) for all \(n,\) can anything be said about \(\Sigma\left(1 / a_{n}\right) ?\) Give reasons for your answer.

The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\}\).

In each of the geometric series, write out the first few terms of the series to find \(a\) and \(r\), and find the sum of the series. Then express the inequality \(|r|<1\) in terms of \(x\) and find the values of \(x\) for which the inequality holds and the series converges. $$\sum_{n=0}^{\infty} 3\left(\frac{x-1}{2}\right)^{n}$$

What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer.

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