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Chapter 7: Problem 45

Find the sum of the convergent series. $$ \sum_{n=1}^{\infty}(\sin 1)^{n} $$

Short Answer

Expert verified
The sum of the convergent series \(\sum_{n=1}^{\infty}(\sin 1)^{n}\) is \(\sin1 / (1 - \sin1)\)

Step by step solution

01

Identify geometric series

The infinite series can be classified as a geometric series where the first term 'a' is \(\sin1\) and the common ratio 'r' is also \(\sin1\). This is since each term is the previous term multiplied by \(\sin1\).
02

Apply sum formula of geometric series

For a geometric series with first term \(a\) and common ratio \(r\) where \(|r| < 1\), the sum \(S\) is given by \(S = a/(1-r)\). Here, \(a = \sin1\) and \(r = \sin1\). Therefore, applying the formula, the sum \(S\) of this convergent series is \(S = \sin1 / (1 - \sin1)\).

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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 sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. This type of series resembles a simple pattern, like a chain reaction, where each event influences the next through consistent multiplication.

When we look at the example of the series \(\sum_{n=1}^{\infty}(\sin 1)^{n}\), we recognize that it's a geometric series. Each term in the series can be obtained by multiplying \(\sin 1\) by the previous term. Therefore, this repeating multiplication by \(\sin 1\) reflects the concept of a common ratio in geometric progressions.
Sum Formula of Geometric Series
Calculating the sum of a geometric series, especially when it's infinite, might seem daunting, but it is made simpler by using the sum formula. This formula is applicable when the absolute value of the common ratio is less than one \(|r| < 1\).

The sum formula of a geometric series is \(S = \frac{a}{1 - r}\), where \(a\) represents the first term and \(r\) the common ratio. Using the given series \(\sum_{n=1}^{\infty}(\sin 1)^{n}\) as an example, the first term \(a\) and the common ratio \(r\) are both \(\sin 1\). Given \(\sin 1\) is less than one, we are assured this sum formula will lead us to the correct answer.
Infinite Series
An infinite series is a summation of an endless sequence of terms. Not all infinite series converge, which means not all have a finite sum. For a geometric series to be convergent, its common ratio must have an absolute value of less than one \(|r| < 1\).

The concept of 'convergence' is pivotal in understanding infinite series. When the terms of an infinite series get progressively smaller and converge to a certain value as they progress, the series has a finite limit. In the case of our example with \(\sin 1\), since this value is between -1 and 1, the series converges and we can find a sum.
Common Ratio
The common ratio is the factor by which consecutive terms in a geometric series are multiplied to obtain the next term. It's not only essential for defining the series itself but also for determining whether the series is convergent or divergent when it's infinite.

For an infinite series to converge, the common ratio must satisfy the condition \(|r| < 1\). This ensures that each term is smaller than the one before it, driving the series towards a finite limit. In the exercise, the common ratio is \(\sin 1\), which meets this condition because the sine function always outputs a result between -1 and 1 for any real number input.

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Most popular questions from this chapter

The random variable \(\boldsymbol{n}\) represents the number of units of a product sold per day in a store. The probability distribution of \(n\) is given by \(P(n) .\) Find the probability that two units are sold in a given day \([P(2)]\) and show that \(P(1)+P(2)+P(3)+\cdots=1\). $$ P(n)=\frac{1}{3}\left(\frac{2}{3}\right)^{n} $$

Prove that if \(\left\\{s_{n}\right\\}\) converges to \(L\) and \(L>0,\) then there exists a number \(N\) such that \(s_{n}>0\) for \(n>N\).

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(n>1\), then \(n !=n(n-1) !\)

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right) $$

Find the sum of the convergent series. $$ 3-1+\frac{1}{3}-\frac{1}{9}+\cdots $$
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