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Magazine Chapter 10: Problem 9
Write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. $$\sum_{n=1}^{\infty}\left(1-\frac{7}{4^{n}}\right)$$
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Identify the General Term
The series is given as \( \sum_{n=1}^{\infty}\left(1-\frac{7}{4^{n}}\right) \). This can be expanded to individual terms where each term follows the pattern \( 1-\frac{7}{4^{n}} \) for each consecutive \( n \).
02
Calculate the First Eight Terms
Substitute values from \( n=1 \) to \( n=8 \) to find the first eight terms:- \( n=1 \): \( 1 - \frac{7}{4} = \frac{4}{4} - \frac{7}{4} = -\frac{3}{4} \)- \( n=2 \): \( 1 - \frac{7}{16} = \frac{16}{16} - \frac{7}{16} = \frac{9}{16} \)- \( n=3 \): \( 1 - \frac{7}{64} = \frac{64}{64} - \frac{7}{64} = \frac{57}{64} \)- \( n=4 \): \( 1 - \frac{7}{256} = \frac{256}{256} - \frac{7}{256} = \frac{249}{256} \)- \( n=5 \): \( 1 - \frac{7}{1024} = \frac{1024}{1024} - \frac{7}{1024} = \frac{1017}{1024} \)- \( n=6 \): \( 1 - \frac{7}{4096} = \frac{4096}{4096} - \frac{7}{4096} = \frac{4089}{4096} \)- \( n=7 \): \( 1 - \frac{7}{16384} = \frac{16384}{16384} - \frac{7}{16384} = \frac{16377}{16384} \)- \( n=8 \): \( 1 - \frac{7}{65536} = \frac{65536}{65536} - \frac{7}{65536} = \frac{65529}{65536} \).The first eight terms of the sequence are \(-\frac{3}{4}, \frac{9}{16}, \frac{57}{64}, \frac{249}{256}, \frac{1017}{1024}, \frac{4089}{4096}, \frac{16377}{16384}, \frac{65529}{65536}\).
03
Determine the Nature of the Series
The series consists of a constant 1 and a geometric series with the common ratio \( r = \frac{1}{4} \), which is less than 1 in absolute value. This indicates that the series within \( \sum_{n=1}^{\infty} -\frac{7}{4^n} \) converges.
04
Find the Sum of the Geometric Series
For a geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum is \( \frac{a}{1-r} \) if \( |r| < 1 \). Here, \( a = -\frac{7}{4} \) and \( r = \frac{1}{4} \). The sum is given by:\[ S = \frac{-\frac{7}{4}}{1 - \frac{1}{4}} = \frac{-\frac{7}{4}}{\frac{3}{4}} = -\frac{7}{3}. \]
05
Find the Total Sum of the Series
The overall series \( \sum_{n=1}^{\infty}\left(1-\frac{7}{4^{n}}\right) \) can be seen as \( \sum_{n=1}^{\infty} 1 - \sum_{n=1}^{\infty} \frac{7}{4^n} \). The first series diverges since \( n \to \infty \). Thus, the series diverges.Overall, the nature of the series is determined by the constant term and the geometric series, so the entire series diverges.
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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. In the context of the problem given, the series is expressed as \( 1 - \frac{7}{4^n} \), showcasing a constant term \(1\) alongside the geometric part \(\frac{7}{4^n}\). Here, the geometric series is represented as \(-\frac{7}{4^n}\), where the first term \(a\) is \(-\frac{7}{4}\) and the common ratio \(r\) is \(\frac{1}{4}\).
To be a geometric series, the ratio \(r\) should be consistent across consecutive terms. For instance:
To be a geometric series, the ratio \(r\) should be consistent across consecutive terms. For instance:
- Term 2: \( 1 - \frac{7}{4^2} = \frac{9}{16} \)
- Term 3: \( 1 - \frac{7}{4^3} = \frac{57}{64} \)
Convergence and Divergence
When determining whether a series converges or diverges, you are essentially investigating if the series approaches a specific value as the number of terms grows indefinitely. A crucial condition for the convergence of a geometric series is that the absolute value of the ratio, \(|r|\), must be less than \(1\).
In our exercise, the geometric portion \(\sum_{n=1}^{\infty} -\frac{7}{4^n}\) demonstrates convergence. The given common ratio for this series is \(r = \frac{1}{4}\), and since \(|r| < 1\), the geometric component converges.
However, the rest of the series includes a constant value \(1\) across all terms, which adds indefinitely as \(n\) increases. Therefore:
In our exercise, the geometric portion \(\sum_{n=1}^{\infty} -\frac{7}{4^n}\) demonstrates convergence. The given common ratio for this series is \(r = \frac{1}{4}\), and since \(|r| < 1\), the geometric component converges.
However, the rest of the series includes a constant value \(1\) across all terms, which adds indefinitely as \(n\) increases. Therefore:
- Convergent: The geometric part \( -\frac{7}{4^n} \) converges independently.
- Divergent: The constant term \(1\) leads the overall series to diverge as it consistently adds \(1\) in perpetuity.
Sum of Series
The sum of a geometric series can be calculated using the formula \( S = \frac{a}{1-r} \) where \(a\) is the initial term, and \(r\) is the common ratio. This formula only works if \(|r| < 1\), indicating the series is convergent.
For the geometric series part of our given problem:
Nevertheless, since the non-geometric term \(1\) results in overall divergence due to its indefinite addition, no finite sum exists for the complete series \(\sum_{n=1}^{\infty} (1 - \frac{7}{4^n})\). Although we can compute the sum of its convergent element, it's important to understand the implications when combined with divergent components.
For the geometric series part of our given problem:
- \(a = -\frac{7}{4}\)
- \(r = \frac{1}{4}\)
Nevertheless, since the non-geometric term \(1\) results in overall divergence due to its indefinite addition, no finite sum exists for the complete series \(\sum_{n=1}^{\infty} (1 - \frac{7}{4^n})\). Although we can compute the sum of its convergent element, it's important to understand the implications when combined with divergent components.
