/*! 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 22 Evaluate each geometric series o... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each geometric series or state that it diverges. $$1+\frac{2}{7}+\frac{2^{2}}{7^{2}}+\frac{2^{3}}{7^{3}}+\cdots$$

Short Answer

Expert verified
Answer: The sum of the given infinite geometric series is 7/5.

Step by step solution

01

Determine the first term (a1) and the common ratio (r)

We can see that the first term in the given series is \(a_1=1\). To find the common ratio, we can divide the second term by the first term. In this case, $$r=\frac{\frac{2}{7}}{1}=\frac{2}{7}$$
02

Check for convergence

A geometric series converges if the absolute value of the common ratio \(|r|\) is less than 1. In this case, we have $$|r|=|\frac{2}{7}|=\frac{2}{7}<1$$ Since the common ratio is less than 1, the given series converges.
03

Calculate the sum of the series

Since we know the geometric series converges, we can use the formula for the sum of an infinite geometric series. $$S=\frac{a_1}{1-r}=\frac{1}{1-\frac{2}{7}}=\frac{1}{\frac{5}{7}}=\frac{7}{5}$$ So, the sum of the given geometric series is \(\frac{7}{5}\).

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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 series in which each term is obtained by multiplying the previous term by a fixed, non-zero number called the "common ratio," denoted as \( r \). The series takes the form:
  • \( a + ar + ar^2 + ar^3 + \, \ldots \)
where \( a \) is the first term of the series.

In a geometric series, the sequence of partial sums can be calculated if the series converges. There are two kinds of geometric series:
  • **Finite Geometric Series:** A series with a finite number of terms.
  • **Infinite Geometric Series:** A series with an infinite number of terms.
In our original exercise, we are dealing with an infinite geometric series where the first term \( a_1 = 1 \), and the common ratio \( r = \frac{2}{7} \). This series pattern involves multiplying each term by \( \frac{2}{7} \) to get the next term.
Convergence
Convergence is a key concept in calculus, especially when dealing with series. For a series to converge, the sum of its terms must approach a finite number as more terms are added. For a geometric series, convergence depends on the common ratio \( r \).

The criterion for convergence of an infinite geometric series is:
  • If \( |r| < 1 \), the series converges.
  • If \( |r| \geq 1 \), the series diverges.
In the provided series, the common ratio \( r = \frac{2}{7} \), and clearly, \( |\frac{2}{7}| = \frac{2}{7} < 1 \). Thus, the series converges. This means as we keep adding more terms of this series, the total sum will approach a specific number.

Understanding convergence helps in determining whether the series has a finite sum or not. In this exercise, because \( r \) is less than 1, the series converges to a sum which can be calculated.
Infinite Series
An infinite series is a sum of infinite terms, and determining its sum depends heavily on whether it converges or diverges. When we discuss infinite series, we often talk about the sequence of partial sums, which allows us to see if the series approaches a limit.

For an infinite geometric series that converges, its sum can be found using the formula:\[S = \frac{a}{1 - r},\]where:
  • \( S \) is the sum of the series.
  • \( a \) is the first term.
  • \( r \) is the common ratio.
In our example, using the formula:
  • First term, \( a = 1 \).
  • Common ratio, \( r = \frac{2}{7} \).
This leads to the sum:\[S = \frac{1}{1 - \frac{2}{7}} = \frac{1}{\frac{5}{7}} = \frac{7}{5}.\]Thus, the infinite series adds up to \( \frac{7}{5} \), showing that even infinite series can have finite sums.

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Most popular questions from this chapter

Marie takes out a \(\$ 20,000\) loan for a new car. The loan has an annual interest rate of \(6 \%\) or, equivalently, a monthly interest rate of \(0.5 \% .\) Each month, the bank adds interest to the loan balance (the interest is always \(0.5 \%\) of the current balance), and then Marie makes a \(\$ 200\) payment to reduce the loan balance. Let \(B_{n}\) be the loan balance immediately after the \(n\) th payment, where \(B_{0}=\$ 20,000\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. Determine how many months are needed to reduce the loan balance to zero.

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)},$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). a. Explain why \(\left\\{F_{n}\right\\}\) is a decreasing sequence. b. Plot \(\left\\{F_{n}\right\\},\) for \(n=1,2, \ldots, 20\). c. Based on your experiments, make a conjecture about \(\lim _{n \rightarrow \infty} F_{n}\).

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\int_{1}^{n} x^{-2} d x$$

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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