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$$\text {Evaluate each series or state that it diverges.}$$ $$\sum_{k=1}^{\infty} \frac{(-2)^{k}}{3^{k+1}}$$

Short Answer

Expert verified
If it converges, evaluate its sum. Answer: The series $$\sum_{k=1}^{\infty} \frac{(-2)^{k}}{3^{k+1}}$$ diverges, so we cannot evaluate its sum.

Step by step solution

01

Write down the general term

The general term of the series is given by: $$a_k = \frac{(-2)^{k}}{3^{k+1}}$$
02

Apply the ratio test

Let's calculate the ratio of consecutive terms, namely \(\frac{a_{k+1}}{a_k}\), and then check the limit as k approaches infinity: $$\frac{a_{k+1}}{a_k} = \frac{\frac{(-2)^{k+1}}{3^{(k+1)+1}}}{\frac{(-2)^{k}}{3^{k+1}}} = \frac{(-2)^{k+1}3^{k+1}}{(-2)^{k}3^{(k+1)+1}}$$ Simplify the expression: $$\frac{a_{k+1}}{a_k} = \frac{(-2)^{k+1}3^{k+1}(3)}{(-2)^{k}3^{(k+1)+1}(-2)} = \frac{(-2)(3)}{-2}$$ The ratio of consecutive terms is a constant that equals \(\frac{(-2)(3)}{-2} = 3\).
03

Determine convergence or divergence

Since the ratio of consecutive terms is constant and equal to 3, we have: $$\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = 3$$ Since the limit is greater than 1, we can conclude that the series diverges according to the ratio test.
04

Provide the final answer

The series $$\sum_{k=1}^{\infty} \frac{(-2)^{k}}{3^{k+1}}$$ diverges, so we cannot evaluate its sum.

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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 crucial tool for determining whether an infinite series converges or diverges. It requires calculating the limit of the absolute value of the ratio of consecutive terms, \( \frac{|a_{k+1}|}{|a_k|} \).

If this limit is less than 1, the series converges; if it's greater than 1, the series diverges; and if it equals 1, the test is inconclusive. It is particularly useful for series whose terms involve factorials, exponentials, or powers of k.

For example, in the exercise \( \frac{a_{k+1}}{a_k} = 3 \), since 3 is greater than 1, we can immediately determine that the series diverges by the ratio test without needing to investigate further. However, if additional context suggests a convergence at the limit of 1, additional methods would be required to ascertain the series' behavior.
Infinite Series
An infinite series is the sum of an infinite sequence of numbers, written in the form \( \sum_{k=1}^{} a_k \), where \( a_k \), the k-th term, is a function of k. This concept is fundamental in many areas of mathematics, including analysis and calculus.

There are different types of series, such as arithmetic, geometric, and harmonic, and each has specific properties that can determine their convergence or divergence. For instance, the series in the exercise is a form of a geometric series, which is a series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.
Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant called the ratio. The general form of a geometric series is \( \sum_{k=0}^{} ar^k \), where 'a' is the first term, and 'r' is the common ratio.

The convergence of a geometric series depends solely on the ratio 'r'. If |r|<1, the series converges; otherwise, it diverges. The series in the given exercise has a ratio of -2/3, and when we apply the ratio test, we find that the series diverges because the absolute value of the common ratio is greater than 1. This indicates that over time, the terms of the series do not approach zero, which is a necessary condition for the convergence of an infinite series.
Limits
The concept of a limit in mathematics describes the value that a function or sequence 'approaches' as the input or index approaches some value. Limits are essential to calculus and mathematical analysis and deal with the behavior of sequences and functions near specific points or at infinity.

In the context of series, the limit of the ratio of consecutive terms, as applied in the ratio test, helps us to understand the behavior of the series at infinity. If the terms of the series grow without bound, the series is said to diverge. Conversely, if the terms approach a finite limit, and in particular, if they approach zero, the series might converge, provided additional conditions are met.

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Most popular questions from this chapter

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=4 a_{n}\left(1-a_{n}\right) ; a_{0}=0.5$$

Consider the alternating series $$ \sum_{k=1}^{\infty}(-1)^{k+1} a_{k}, \text { where } a_{k}=\left\\{\begin{array}{cl} \frac{4}{k+1}, & \text { if } k \text { is odd } \\ \frac{2}{k}, & \text { if } k \text { is even } \end{array}\right. $$ a. Write out the first ten terms of the series, group them in pairs, and show that the even partial sums of the series form the (divergent) harmonic series. b. Show that \(\lim _{k \rightarrow \infty} a_{k}=0\) c. Explain why the series diverges even though the terms of the series approach zero.

A fallacy Explain the fallacy in the following argument. Let \(x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots \cdot\) It follows that \(2 y=x+y\) which implies that \(x=y .\) On the other hand, $$ x-y=\underbrace{\left(1-\frac{1}{2}\right)}_{>0}+\underbrace{\left(\frac{1}{3}-\frac{1}{4}\right)}_{>0}+\underbrace{\left(\frac{1}{5}-\frac{1}{6}\right)}_{>0}+\cdots>0 $$ is a sum of positive terms, so \(x>y .\) Therefore, we have shown that \(x=y\) and \(x>y\)

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