Problem 45
Question
Find the sum of the convergent series. $$ \sum_{n=1}^{\infty}(\sin 1)^{n} $$
Step-by-Step Solution
Verified Answer
The sum of the convergent series \(\sum_{n=1}^{\infty}(\sin 1)^{n}\) is \(\sin1 / (1 - \sin1)\)
1Step 1: 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 \(\sin1\).
2Step 2: 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 \(S = \sin1 / (1 - \sin1)\).
Key Concepts
Geometric SeriesSum Formula of Geometric SeriesInfinite SeriesCommon Ratio
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.
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.
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.
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.
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.
Other exercises in this chapter
Problem 44
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