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Chapter 8: Problem 37

Evaluate each geometric series or state that it diverges. $$3 \sum_{k=0}^{\infty}(-\pi)^{-k}$$

Short Answer

Expert verified
Answer: The sum of the given geometric series is $$\frac{3\pi}{\pi+1}$$.

Step by step solution

01

Check convergence/divergence

First, check if the geometric series converges by determining if the absolute value of the common ratio is less than 1: $$|-1/\pi| = 1/\pi < 1$$ Since \(1/\pi < 1\), the geometric series converges.
02

Find the first term

The first term of the series is when \(k=0\), so we have: $$a = 3(-\pi)^{-0} = 3(1) = 3$$
03

Calculate the sum of the geometric series

The sum of the convergent geometric series can be found using the formula \(S = \frac{a}{1-r}\). In our case, \(a = 3\) and \(r = -1/\pi\): $$S = \frac{3}{1-(-1/\pi)} = \frac{3}{1+1/\pi} = \frac{3}{\frac{\pi+1}{\pi}} = \frac{3\pi}{\pi+1}$$ The sum of the given geometric series is: $$3 \sum_{k=0}^{\infty}(-\pi)^{-k} = \frac{3\pi}{\pi+1}$$

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Key Concepts

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

Convergence of Series
Understanding when a series converges is crucial for calculating its sum. Convergence means that as we add more and more terms of the series, the total sum approaches a finite value.

To determine convergence for a geometric series, we must look at the series's common ratio, which is the constant multiplier between consecutive terms. If the absolute value of the common ratio is less than 1, the series will converge.

In our example, the common ratio is \(-1/\pi\), and its absolute value is \(1/\pi\), which is less than 1. Therefore, we can confirm that the series converges. Convergence is essential since it indicates that a finite sum exists for the entire series.
Common Ratio
The common ratio is a term that appears frequently when dealing with geometric series. It's the ratio that we multiply by each term of the series to get the next term. In simpler terms, it's what makes a series geometric.

For the series \(3 \sum_{k=0}^{\infty}(-\pi)^{-k}\), we find the common ratio by looking at how the terms change from one to the next. In this case, the common ratio is \(-1/\pi\). The common ratio is key to determining both the convergence of the series and its sum.
Sum of a Series
Once we've established a geometric series converges, we can calculate its sum. The formula to add up all the terms of a convergent geometric series is given as \(S = \frac{a}{1-r}\), where \(S\) represents the sum, \(a\) is the first term, and \(r\) is the common ratio.

In our exercise, the first term \(a\) is 3, and the common ratio \(r\) is \(-1/\pi\). Plugging these into our formula gives us \(S = \frac{3}{1 + 1/\pi}\), which simplifies to \(\frac{3\pi}{\pi+1}\). This sum represents the limit that the series approaches as the number of terms grows infinitely large.
Infinite Series
An infinite series is a sum of an infinite sequence of numbers. While the idea of adding up an infinite number of terms might seem perplexing, many infinite series can actually sum to a finite number, such as the geometric series we discussed.

The concept of an infinite series demands a strong understanding of limits since we're essentially looking for the value that the series is approaching rather than adding up an actual infinite list of terms. In our example, even though the series involves an infinite number of terms, it converges to the finite value of \(\frac{3\pi}{\pi+1}\).

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Most popular questions from this chapter

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b>1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n !>b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) with terms that are non increasing in magnitude, converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n} .\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k .\) It can be shown that for \(n \geq 1\) \(\left|S-\left(S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right)\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|\) a. Interpret this inequality and explain why it is a better approximation to \(S\) than \(S_{n}\) b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

For a positive real number \(p,\) the tower of exponents \(p^{p^{p}}\) continues indefinitely and the expression is ambiguous. The tower could be built from the top as the limit of the sequence \(\left\\{p^{p},\left(p^{p}\right)^{p},\left(\left(p^{p}\right)^{p}\right)^{p}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{1}=p^{p} .\) The tower could also be built from the bottom as the limit of the sequence \(\left\\{p^{p}, p^{\left(p^{p}\right)}, p^{\left(p^{(i)}\right)}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=p^{a_{n}}(\text { building from the bottom })\) where again \(a_{1}=p^{p}\). a. Estimate the value of the tower with \(p=0.5\) by building from the top. That is, use tables to estimate the limit of the sequence defined recursively by (1) with \(p=0.5 .\) Estimate the maximum value of \(p > 0\) for which the sequence has a limit. b. Estimate the value of the tower with \(p=1.2\) by building from the bottom. That is, use tables to estimate the limit of the sequence defined recursively by (2) with \(p=1.2 .\) Estimate the maximum value of \(p > 1\) for which the sequence has a limit.

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 \(10 ?\)
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