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Chapter 8: Problem 19

Evaluate each geometric series or state that it diverges. $$\sum_{k=0}^{\infty}\left(\frac{1}{4}\right)^{k}$$

Short Answer

Expert verified
Answer: The sum of the given infinite geometric series is \(\frac{4}{3}\).

Step by step solution

01

Find the common ratio of the series

The given series is a geometric series in the form: $$\sum_{k=0}^{\infty}\left(\frac{1}{4}\right)^{k}$$ Here, the common ratio, denoted as "r", is: $$r = \frac{1}{4}$$
02

Determine if the series converges

A geometric series converges if and only if the common ratio (r) is between -1 and 1 (excluding -1 and 1). In this case, we have: $$r = \frac{1}{4}$$ Since \(\frac{1}{4}\) is between -1 and 1, the series converges.
03

Use the geometric series sum formula to evaluate the sum

Now that we know the series converges, we can use the geometric series sum formula to evaluate the sum. The formula is: $$S_{\infty} = \frac{a}{1 - r}$$ Where \(S_{\infty}\) is the sum of the converging series, \(a\) is the first term of the series, and \(r\) is the common ratio. In this case, the first term of the series is: $$a = \left(\frac{1}{4}\right)^0 = 1$$ So the sum of the series is: $$S_{\infty} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$ Therefore, the sum of the given geometric series is: $$\sum_{k=0}^{\infty}\left(\frac{1}{4}\right)^{k} = \frac{4}{3}$$

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Key Concepts

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

Convergence of Series
When working with series, convergence is a key concept to understand. Convergence tells us whether the series approaches a certain value as more terms are added. For a series to converge, the terms must get smaller and approach zero. For a **geometric series**, specifically, convergence depends on the **common ratio**.
  • If the common ratio \( r \) is between \(-1\) and \(1\) (excluding both), the series will converge.
  • If \(|r| \geq 1\), the geometric series will diverge, meaning it won't approach a single value.
In the example given, the common ratio is \( \frac{1}{4} \), which lies within the range for convergence. Hence, the series converges.
Geometric Series Formula
The geometric series formula is a powerful tool for finding the sum of an infinite geometric series. A geometric series is one where each term is a constant multiple, called the common ratio, \( r \), of the previous term. The general form of such a series is: \[ a + ar + ar^2 + ar^3 + \cdots \]For a series with an infinite number of terms that converges, the sum \( S_\infty \) is given by the formula:\[ S_\infty = \frac{a}{1 - r} \]
  • \( a \) is the first term of the series.
  • \( r \) is the common ratio.
In our example, the series starts at 1 (since \((\frac{1}{4})^0 = 1\)) and has a ratio of \( \frac{1}{4} \). Plugging these into the formula, we find the sum to be \( \frac{4}{3} \).
Infinite Series
Infinite series are series that continue indefinitely without terminating. The sum of an infinite series is typically denoted as \( \sum_{k=0}^{\infty} a_k \). These can be challenging to understand because they involve summing an endless number of terms. For some infinite series, the total sum approaches a fixed value鈥攅ven though the series continues infinitely. These are the ones that converge. The ability to calculate a specific number for their sum arises from their **convergence**. This is unlike divergent series, where the terms add up to infinity or fail to approach any limiting value.The given series is a classic example of a converging infinite geometric series, starting from \( k=0 \) to \( \infty \) with a common ratio of \( \frac{1}{4} \). Understanding whether an infinite series converges or diverges is essential for calculating its sum effectively.

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Most popular questions from this chapter

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. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100\)

Prove that if \(\sum a_{k}\) diverges, then \(\sum c a_{k}\) also diverges, where \(c \neq 0\) is a constant.

The Greek philosopher Zeno of Elea (who lived about 450 B.c.) invented many paradoxes, the most famous of which tells of a race between the swift warrior Achilles and a tortoise. Zeno argued The slower when running will never be overtaken by the quicker: for that which is pursuing must first reach the point from which that which is fleeing started, so that the slower must necessarily always be some distance ahead. In other words, by giving the tortoise a head start, Achilles will never overtake the tortoise because every time Achilles reaches the point where the tortoise was, the tortoise has moved ahead. Resolve this paradox by assuming that Achilles gives the tortoise a 1 -mi head start and runs \(5 \mathrm{mi} / \mathrm{hr}\) to the tortoise's \(1 \mathrm{mi} / \mathrm{hr}\). How far does Achilles run before he overtakes the tortoise, and how long does it take?

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 \(b=10\).

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.
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