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Evaluate the series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) two ways. a. Use a telescoping series argument. b. Use a geometric series argument after first simplifying \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\)

Short Answer

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Question: Evaluate the infinite arithmetic series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\). Answer: \(\frac{1}{2}\).

Step by step solution

01

Identify the telescoping series

A telescoping series is a series in which each term can be expressed as the difference of two consecutive terms of a sequence. In this case, the given series can be reconstructed as the sum of differences between consecutive terms of a divergent series.
02

Rewrite the series as a sum of differences

To see the telescoping nature of the series, we need to rewrite our series as a sum of differences between consecutive terms of the sequence \(\left\{ \frac{1}{2^k}\right\}\). $$ \sum_{k=1}^{\infty} \left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)= \left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\cdots $$
03

Cancel terms and identify remaining terms

In the rewritten series, most of the terms will cancel. Here's how it looks: $$ \sum_{k=1}^{\infty} \left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)=\frac{1}{2}-\cancel{\frac{1}{4}}+\cancel{\frac{1}{4}}-\cancel{\frac{1}{8}}+\cancel{\frac{1}{8}}-\cdots $$ The terms cancelling out is what gives a telescoping series its name. In our case, the only remaining term is the first term: \(\frac{1}{2}\).
04

Sum the remaining terms

Since the only remaining term is \(\frac{1}{2}\), the sum of the series is: $$ \sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)=\frac{1}{2} $$ b. Geometric series argument
05

Simplify the expression inside the sum

First, we simplify the given expression \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\). To do this, we factor out \(2^{-k}\): $$ \frac{1}{2^{k}}-\frac{1}{2^{k+1}}=2^{-k}\left(1-\frac{1}{2}\right)=2^{-k}\left(\frac{1}{2}\right)=\frac{1}{2^{k+1}} $$
06

Recognize the new series as a geometric series

Now, our series can be rewritten as: $$ \sum_{k=1}^{\infty}\frac{1}{2^{k+1}} $$ which is a geometric series with the common ratio \(r=\frac{1}{2}\) and the first term \(a=\frac{1}{2^2}=\frac{1}{4}\).
07

Use the formula for the sum of an infinite geometric series

To find the sum of an infinite geometric series, we use the formula: $$ S=\frac{a}{1-r} $$ In our case, \(a=\frac{1}{4}\) and \(r=\frac{1}{2}\).
08

Compute the sum

Now we can compute the sum of the series: $$ S=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2} $$ So the sum of the series is \(\frac{1}{2}\), using both a telescoping series argument and a geometric series argument after simplifying the expression.

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Key Concepts

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

Geometric Series
When studying sequences and series, a geometric series is one of the most fundamental concepts. It is characterized by a constant ratio between successive terms. Specifically, a series is considered geometric if each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. The geometric series can be written as:

\[ S = a + ar + ar^2 + ar^3 + \dotsb + ar^{n-1} \]
where \( a \) is the first term, and \( r \) is the common ratio. A fascinating aspect of a geometric series is that if \( |r| < 1 \) (the absolute value of the common ratio is less than one), the series has a finite sum, even if it has infinitely many terms. This sum is found using the following formula:

\[ S = \frac{a}{1-r} \]
Understanding the behavior of geometric series is crucial for various mathematical applications, from calculating loan repayments to understanding natural phenomena. The simplicity of its pattern makes it a powerful tool in both theoretical and practical scenarios. Additionally, examining infinite series like these opens the door to understanding convergence and the value of a series in the context of limits.
Convergent Series
When diving into the world of infinite series, encountering a convergent series is inevitable. An infinite series is simply the sum of an infinite sequence of numbers. The key question is whether this sum approaches a finite limit or not. A series is said to be convergent if it approaches some finite value as more and more terms are added. In other words, the partial sums of the series form a sequence that gets arbitrarily close to a particular number, known as the limit.

To put it into perspective, let's say you’re adding fractions of a whole continuously, and as you add more fractions, the total sum you get is coming closer to a certain number without ever exceeding it. That's what convergence in series is all about. Demonstrating that an infinite series is convergent is essential because it provides assurance that working with its sum makes sense and won't lead to infinite or undefined results. Convergence is a central topic in calculus and analysis because it lays the foundation for integrating and differentiating power series, understanding the behavior of functions, and more.
Infinite Series
In calculus and mathematical analysis, an infinite series is a sum of infinitely many terms. This can sound rather peculiar at first, as it stretches the concept of 'adding things up' into the realm of the infinite. But with the proper mathematical framework, infinite series become not only manageable but also incredibly useful. Examples include power series, Fourier series, and yes, the geometric series.

However, not all infinite series have a finite sum. Determining whether a series is convergent or divergent (where terms don't tend toward a finite limit) is vital in understanding its behavior. It's like an ongoing journey — if the path you’re taking will eventually lead you to a particular destination, your journey (series) is convergent. But if you keep walking forever with no end in sight, your path is divergent.

Infinite series are not just academic exercises; they have practical applications in solving differential equations, modeling natural phenomena, and in the fields of engineering and physics. Whenever the sum of an infinite series is finite, it’s a powerful statement about the harmony of infinitude within the bounds of mathematics.

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Most popular questions from this chapter

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0},\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G.)\) a. Show that \(a_{n} > b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 3^{-k}$$

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{c n}{b n+1}=\frac{c}{b}, \text { for real numbers } b > 0 \text { and } c > 0$$

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{10} \text { and } b_{n}=n^{9} \ln ^{3} n, n \geq 7$$

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. Radioactive decay A material transmutes \(50 \%\) of its mass to another element every 10 years due to radioactive decay. Let \(M_{n}\) be the mass of the radioactive material at the end of the \(n\) th decade, where the initial mass of the material is \(M_{0}=20 \mathrm{g}\)

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