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Chapter 8: Problem 9

Write the first four terms of the sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\) $$a_{n}=1 / 10^{n}$$

Short Answer

Expert verified
Answer: The first four terms of the sequence are \(\frac{1}{10}\), \(\frac{1}{100}\), \(\frac{1}{1000}\), and \(\frac{1}{10000}\).

Step by step solution

01

Set up the Formula

We are given the formula for the sequence as \(a_{n} = \frac{1}{10^n}\). Our task is to find the first four terms, which means we will substitute \(n=1,2,3\), and \(4\) into the formula.
02

Find the First Term (n=1)

To find the first term, we plug \(n=1\) into the formula: $$a_1 = \frac{1}{10^1} = \frac{1}{10}$$
03

Find the Second Term (n=2)

To find the second term, we plug \(n=2\) into the formula: $$a_2 = \frac{1}{10^2} = \frac{1}{100}$$
04

Find the Third Term (n=3)

To find the third term, we plug \(n=3\) into the formula: $$a_3 = \frac{1}{10^3} = \frac{1}{1000}$$
05

Find the Fourth Term (n=4)

To find the fourth term, we plug \(n=4\) into the formula: $$a_4 = \frac{1}{10^4} = \frac{1}{10000}$$ The first four terms of the sequence are \(\frac{1}{10}\), \(\frac{1}{100}\), \(\frac{1}{1000}\), and \(\frac{1}{10000}\).

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Key Concepts

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

Infinite Series
An infinite series is essentially a sum of an infinite list of numbers. This concept is central in calculus and analysis. When we talk about an infinite series, we typically look at the sum of terms of a sequence. Each term in the sequence is added one by one, building on all that have come before it. For example, consider an infinite geometric series, like our sequence \( a_{n} = \frac{1}{10^n} \). We can write the series as:
  • \( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \cdots \)
When evaluating these types of series, questions often arise about convergence. Does adding more terms lead us to a specific number, or does it just grow infinitely? For our particular sequence, the series converges to a finite number (less than 1), as each term gets very small, very quickly.
To determine convergence formally, tests like the ratio or root test in calculus are often used. These powerful tools help decide the behavior of endless sums.
Term Calculation
Calculating terms in a sequence can be straightforward once you understand the rule or formula governing the sequence. In exercises, you're often asked to find specific terms by plugging in values for \(n\), the position in the sequence. Our sequence's formula was \( a_{n} = \frac{1}{10^n} \). Here's how you calculate terms:
  • Identify the term position \(n\) you are solving for.
  • Plug \(n\) into the formula.
  • Compute the result.
For example, to find the third term in our sequence, you'd substitute 3 for \(n\) in \( \frac{1}{10^n} \), yielding \( \frac{1}{1000} \).
Remembering the sequence rule not only helps in calculating the terms but also in predicting how they will behave.
As \(n\) increases, terms in our specific sequence become much smaller, which influences series behavior significantly, as it does with geometric progressions.
Exponential Functions
Exponential functions are functions where a constant base is raised to a variable exponent. They are crucial in understanding growth processes and decay in mathematics, such as compound interest or radioactive decay.
In the context of our sequence, \( a_{n} = \frac{1}{10^n} \), we notice that the denominator involves an exponential function. Here's what makes exponential functions unique:
  • They grow very quickly or decay rapidly, depending on the base.
  • If the base is greater than 1, the function grows; if it is between 0 and 1, like our case, it decays.
The function \( 10^n \) grows as \(n\) increases, making the terms \( \frac{1}{10^n} \) shrink respectively.
Understanding exponentials helps in visualizing the vast range of sequence behaviors, predicting long-term tendencies, and provides a significant tool in calculus, particularly when solving differential equations.

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Most popular questions from this chapter

a. Evaluate the series $$ \sum_{k=1}^{\infty} \frac{3^{k}}{\left(3^{k+1}-1\right)\left(3^{k}-1\right)} $$ b. For what values of \(a\) does the series $$ \sum_{k=1}^{\infty} \frac{a^{k}}{\left(a^{k+1}-1\right)\left(a^{k}-1\right)} $$ converge, and in those cases, what is its value?

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

In \(1978,\) in an effort to reduce population growth, China instituted a policy that allows only one child per family. One unintended consequence has been that, because of a cultural bias toward sons, China now has many more young boys than girls. To solve this problem, some people have suggested replacing the one-child policy with a one-son policy: A family may have children until a boy is born. Suppose that the one-son policy were implemented and that natural birth rates remained the same (half boys and half girls). Using geometric series, compare the total number of children under the two policies.

The fractal called the snowflake island (or Koch island ) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length \(1 .\) The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) with a new outward equilateral triangle with sides of length \(1 / 3\) (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) with a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n \rightarrow \infty\) is called the snowflake island. a. Let \(L_{n}\) be the perimeter of \(I_{n} .\) Show that \(\lim _{n \rightarrow \infty} L_{n}=\infty\) b. Let \(A_{n}\) be the area of \(I_{n} .\) Find \(\lim _{n \rightarrow \infty} A_{n} .\) It exists!

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?
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