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Chapter 8: Problem 40

Evaluate each geometric series or state that it diverges. $$\sum_{k=1}^{\infty} 3\left(-\frac{1}{8}\right)^{3 k}$$

Short Answer

Expert verified
If it converges, what is its sum? Answer: The given geometric series converges, and its sum is $$-\frac{3}{511}.$$

Step by step solution

01

Identify the first term and the common ratio

First, let's identify the key components of the given geometric series: $$\sum_{k=1}^{\infty} 3\left(-\frac{1}{8}\right)^{3 k}$$ The first term of the series is when k = 1: $$3 \left(-\frac{1}{8}\right)^{3(1)} = 3 \left(-\frac{1}{8}\right)^3$$ So, the first term is: $$a = 3\left(-\frac{1}{8}\right)^3$$ The common ratio is clearly the expression inside the parentheses raised to the power of 3k: $$r = \left(-\frac{1}{8}\right)^3.$$
02

Determine convergence or divergence

The geometric series converges if |r| < 1. Let's check if this condition applies: $$\left|\left(-\frac{1}{8}\right)^3\right| = \frac{1}{8^3} = \frac{1}{512}$$ Since \(\frac{1}{512} < 1\), this series converges.
03

Evaluate the converging geometric series

Since this geometric series converges, we can now use the formula for the sum of a converging geometric series: $$S_\infty = \frac{a}{1 - r}$$ Plug in the values for \(a\) and \(r\): $$S_\infty = \frac{3 \left(-\frac{1}{8}\right)^3}{1 - \left(-\frac{1}{8}\right)^3}.$$ Now simplify the expression: $$S_\infty = \frac{3 \left(-\frac{1}{8}\right)^3}{1 - \frac{1}{512}} = \frac{-\frac{3}{512}}{\frac{511}{512}}$$ Finally, perform the division: $$S_\infty = - \frac{3}{511}$$ Thus, the geometric series converges, and its sum is: $$\sum_{k=1}^{\infty} 3\left(-\frac{1}{8}\right)^{3 k} = -\frac{3}{511}.$$

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Key Concepts

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

Convergence
In mathematics, convergence is an essential concept when dealing with series, particularly infinite series. It indicates whether the series settles towards a specific value as the number of terms grows indefinitely. A geometric series converges if its common ratio, denoted as \( r \), satisfies the condition \( |r| < 1 \). This condition ensures the terms keep decreasing in magnitude, allowing the series to sum up to a finite value.

For the geometric series \( \sum_{k=1}^{\infty} 3\left(-\frac{1}{8}\right)^{3 k} \), the common ratio is \( r = \left(-\frac{1}{8}\right)^3 \), and its absolute value is \( \frac{1}{512} \), which is indeed less than 1. Therefore, the series converges, and we can find its sum using the sum formula for infinite geometric series.
Common Ratio
The common ratio is a vital element in any geometric series and fundamentally dictates the series' behavior. In a geometric series, each term is obtained by multiplying the previous term by a fixed, not-zero number called the common ratio \( r \). As the series progresses, this ratio governs how quickly the terms approach zero or amplify.

In our given series, the common ratio is \( r = \left(-\frac{1}{8}\right)^3 \). When calculating this, the result is \( r = -\frac{1}{512} \). The negative sign indicates alternating positive and negative terms, whereas the small value of \( \frac{1}{512} \) implies that each subsequent term becomes significantly smaller in magnitude. Understanding the common ratio is critical when determining the convergence and sum of the series.
Infinite Series
An infinite series is the sum of an infinite sequence of numbers or terms. While it might seem daunting due to its endless nature, many infinite series can converge to a finite value, as in the case of geometric series.

For a geometric series, identifying whether it sums to a finite value involves checking the absolute value of the common ratio. If \( |r| < 1 \), the series is convergent, meaning the infinite sum approaches a specific number.

In practical terms, the sum of a convergent infinite geometric series is found using the formula \( S_\infty = \frac{a}{1 - r} \), where \( a \) is the first term. Thus, understanding the nature of infinite series allows you to resolve otherwise seemingly unmanageable problems with precision and clarity.

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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. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2$$

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}_{n=2}^{\infty}.$$

Suppose that you take 200 mg of an antibiotic every 6 hr. The half-life of the drug is 6 hr (the time it takes for half of the drug to be eliminated from your blood). Use infinite series to find the long-term (steady-state) amount of antibiotic in your blood.
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