/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Let \(b_{n}\) be an infinite seq... [FREE SOLUTION] | 91影视

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Let \(b_{n}\) be an infinite sequence of zeros and ones. What is the largest possible value of \(x=\sum_{n=1}^{\infty} b_{n} / 2^{n} ?\)

Short Answer

Expert verified
The largest possible value of x is 1.

Step by step solution

01

Understanding the Infinite Series

The sum given in the problem is an infinite series in the form of a binary decimal expansion. Each term in the series contributes to the final value of x based on the position coefficient which is a power of 2.
02

Analyzing Sequence Values

Since each term in the sequence, denoted by \(b_n\), is either 0 or 1, the value of a term \(b_n/2^n\) can be either \(0\) or \(1/2^n\). To maximize the series x, we need to maximize each contributing term.
03

Maximizing Each Term

To maximize each term in the sequence \(b_n/2^n\), we need to set all \(b_n=1\) as long as the series converges. Setting \(b_n=1\) for all \(n\) yields the following series: \[ x = \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]
04

Solving the Infinite Geometric Series

The series \(x = \sum_{n=1}^{\infty} \frac{1}{2^n}\) is a geometric series with the first term \(a=\frac{1}{2}\) and the common ratio \(r=\frac{1}{2}\). The sum of an infinite geometric series is \(\frac{a}{1-r}\). Substituting the values, we get \[ x = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{1}{2 \times \frac{1}{2}} = 1 \].

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Key Concepts

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

Geometric Series
A geometric series is a type of series where each term is a constant multiple, called the "common ratio," of the previous term. In the sequence given in the exercise, each term is of the form \( \frac{1}{2^n} \), where the terms decrease by a factor of \( \frac{1}{2} \). This makes it a classic example of a geometric series.
To compute the sum of an infinite geometric series, we apply the formula: \( \text{Sum} = \frac{a}{1-r} \). Here, \(a\) is the first term of the series, and \(r\) is the common ratio between the terms. While finite geometric series are simply added up like regular sums, infinite ones depend on the behavior as they progress towards infinity. If the common ratio \(r\) has an absolute value less than 1, the series will converge to a finite limit.
In our exercise, with \(a = \frac{1}{2}\) and \(r = \frac{1}{2}\), the series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \) converges to \(1\), meaning the infinite sum is equal to \(1\) when all conditions are satisfied.
Binary Decimal Expansion
Binary decimal expansion is a technique used to express numbers in the binary numeral system, which uses only the digits 0 and 1. Like the decimal system where each digit represents a power of ten, in binary each digit represents a power of two.
The infinite series \( \sum_{n=1}^{\infty} \frac{b_n}{2^n} \) translates a sequence of 0s and 1s into a number between 0 and 1. Each term, \( \frac{b_n}{2^n} \), represents a fractional part, where the coefficient \(b_n\) determines if that fractional part is contributing (1) or not (0).
For example, the series \(1/2 + 1/4 + 1/8 + \ldots\) approximates a binary fraction that would look like 0.111... in binary form, which is equivalent to 1 in the decimal system. Binary decimal expansions are therefore particularly useful for representing real numbers inside of computers, as they inherently work using binary logic.
Convergent Series
A convergent series is a series where the sum of its terms approaches a specific finite number as more terms are added. For a series to convergent, it must have terms that diminish in size fast enough, typically if each term's contribution gets progressively smaller.
In our exercise's infinite series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \), the terms \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \) become smaller and smaller. This rapid decrease helps the series to sum up to a definite value鈥攊n this case, 1. Since the series meets the condition of reducing terms and a common ratio less than 1, it can be concluded that the series converges.
Convergent series are crucial in various mathematical and practical applications because they enable calculations and predictions about behaviors that would otherwise appear to involve infinitely many parts. They provide a foundational element for calculus, analysis, and even modern computation.

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Most popular questions from this chapter

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series \(\sum a_{k}\) with given terms \(a_{k}\) converges, or state if the test is inconclusive. $$ \left.a_{k}=\left(\frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2 k}\right)^{k} \text { (Hint: Compare } a_{k}^{1 / k} \text { to } \int_{k}^{2 k} \frac{d t}{t} .\right) $$

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.If \(b_{n} \geq 0\) is decreasing and \(\sum_{n=1}^{\infty}(-1)^{n-1} b_{n}\) converges conditionally but not absolutely, then \(b_{n}\) does not tend to zero.

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.Suppose that \(a_{n}\) is a sequence such that \(\sum_{n=1}^{\infty} a_{n} b_{n}\) converges for every possible sequence \(b_{n}\) of zeros and ones. Does \(\sum_{n=1}^{\infty} a_{n}\) converge absolutely?

The following alternating series converge to given multiples of \(\pi .\) Find the value of \(N\) predicted by the remainder estimate such that the Nth partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum \(N\) for which the error bound holds, and give the desired approximate value in each case. Up to 15 decimals places, \(\pi=3.141592653589793 .\)[T] \(\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\), error \(<0.0001\)

Suppose that \(\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=p .\) For which values of \(r>0\) is \(\sum_{n=1}^{\infty} r^{n} a_{n}\) guaranteed to converge?

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