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Chapter 8: Problem 34

Evaluate each geometric series or state that it diverges. $$\sum_{k=2}^{\infty}\left(\frac{3}{8}\right)^{3 k}$$

Short Answer

Expert verified
$$\sum_{k=2}^{\infty}\left(\frac{3}{8}\right)^{3 k}$$ Answer: The sum of the given geometric series is \(\frac{27}{485}\).

Step by step solution

01

Identify the common ratio

The given series is: $$\sum_{k=2}^{\infty}\left(\frac{3}{8}\right)^{3 k}$$ The common ratio of this series is the expression inside the sum, which is \(\left(\frac{3}{8}\right)^{3 k}\). In order to work with the sum formula, let us simplify the expression for our next steps.
02

Simplify the common ratio

We can simplify the expression for the common ratio as follows: $$\left(\frac{3}{8}\right)^{3 k} = \left(\left(\frac{3}{8}\right)^3\right)^k = \left(\frac{27}{512}\right)^k$$ The common ratio is now in the form \(ar^k\), with \(a=\frac{27}{512}\) and \(r=\frac{27}{512}\).
03

Check for convergence

A geometric series converges if the absolute value of the common ratio \(|r|\) is less than 1. In this case, we have: $$\left|\frac{27}{512}\right| = \frac{27}{512} < 1$$ Since the common ratio is less than 1, the series converges.
04

Find the sum of the series

Now that we know the series converges, we can use the formula for the sum of an infinite geometric series: $$S_\infty = \frac{a}{1-r}$$ Substitute the values of \(a\) and \(r\) we found earlier: $$S_\infty = \frac{\frac{27}{512}}{1-\frac{27}{512}}$$
05

Simplify the sum

Now we can simplify the expression for the sum of the series: $$S_\infty = \frac{\frac{27}{512}}{\frac{485}{512}} = \frac{27}{485}$$ So, the sum of the given geometric series is \(\frac{27}{485}\).

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Key Concepts

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

Common Ratio
The common ratio is a pivotal concept when dealing with geometric series. It is the ratio between any term and its preceding term within a sequence. In mathematical terms, for a sequence of numbers where each term after the first is found by multiplying the preceding term by a fixed, non-zero number, this multiplier is known as the common ratio, denoted as 'r'.

In the given exercise, the series is represented as \( \sum_{k=2}^{\infty}\left(\frac{3}{8}\right)^{3 k} \) and the common ratio needs to be identified for the series to be evaluated. As highlighted in the solution, by simplifying \( \left(\frac{3}{8}\right)^{3 k} \) to \( \left(\frac{27}{512}\right)^k \) we deduce the common ratio to be \( \frac{27}{512} \). This value is crucial as it determines whether the series converges or diverges, which leads us seamlessly into our next topic.
Infinite Geometric Series
An infinite geometric series is a sequence of numbers in which each term after the first is found by multiplying the previous term by a common ratio, and this sequence continues indefinitely. The convergence or divergence of such a series depends on the common ratio 'r'.

If \( |r| < 1 \), the series converges, meaning its terms approach a fixed sum as more terms are added. Conversely, if \( |r| \geq 1 \), the series diverges, and no sum can be attributed to the infinite sequence. In our example, since \( |\frac{27}{512}| < 1 \), we establish the series converges. Understanding the conditions for convergence is essential for grasping the behavior of infinite geometric series and applying the appropriate formulas to find their sum.
Sum of a Geometric Series
The sum of a geometric series can be found using a specific formula if and only if the series converges. This usually requires that the common ratio, 'r', must have an absolute value less than one. For an infinite geometric series with a convergence condition met, the formula for the sum is given by \( S_\infty = \frac{a}{1-r} \) where 'a' is the first term and 'r' is the common ratio.

In the provided exercise, we substitute the simplified values of 'a' and 'r' into the formula to get \( S_\infty = \frac{\frac{27}{512}}{1-\frac{27}{512}} \), which upon further simplification yields \( S_\infty = \frac{27}{485} \). This step represents the culmination of identifying the common ratio and verifying convergence, by finding the actual sum if the series is indeed convergent.

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Most popular questions from this chapter

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd .} \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7\), and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

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 \(10 ?\)

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. Population growth When a biologist begins a study, a colony of prairie dogs has a population of \(250 .\) Regular measurements reveal that each month the prairie dog population increases by \(3 \%\) Let \(p_{n}\) be the population (rounded to whole numbers) at the end of the \(n\) th month, where the initial population is \(p_{0}=250\).

It can be proved that if a series converges absolutely, then its terms may be summed in any order without changing the value of the series. However, if a series converges conditionally, then the value of the series depends on the order of summation. For example, the (conditionally convergent) alternating harmonic series has the value $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2 $$ Show that by rearranging the terms (so the sign pattern is \(++-\) ), $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2} \ln 2 $$
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