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Chapter 11: Problem 9

Write out the first few terms of each series to show how the series starts. Then find the sum of the series. $$ \sum_{n=1}^{\infty} \frac{7}{4^{n}} $$

Short Answer

Expert verified
The series starts with terms \( \frac{7}{4}, \frac{7}{16}, \frac{7}{64}, \ldots \) and its sum is \( \frac{7}{3} \).

Step by step solution

01

Identify the Common Ratio

The given series is \( \sum_{n=1}^{\infty} \frac{7}{4^n} \). This is a geometric series where the first term is \( \frac{7}{4^1} = \frac{7}{4} \) and the common ratio \( r \) is \( \frac{1}{4} \).
02

Write the First Few Terms

To find the first few terms of the series, substitute \( n = 1, 2, 3, \ldots \) into \( \frac{7}{4^n} \). The terms are:- First term: \( \frac{7}{4^1} = \frac{7}{4} \)- Second term: \( \frac{7}{4^2} = \frac{7}{16} \)- Third term: \( \frac{7}{4^3} = \frac{7}{64} \)- Fourth term: \( \frac{7}{4^4} = \frac{7}{256} \)
03

Use the Geometric Series Formula

The formula for the sum \( S \) of an infinite geometric series \( \sum_{n=0}^{\infty} ar^n \) is \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( |r| < 1 \).
04

Calculate the Sum of the Series

In the series \( \sum_{n=1}^{\infty} \frac{7}{4^n} \):- First term \( a = \frac{7}{4} \)- Common ratio \( r = \frac{1}{4} \)Plug these into the formula: \[ S = \frac{\frac{7}{4}}{1 - \frac{1}{4}} = \frac{\frac{7}{4}}{\frac{3}{4}} = \frac{7}{3} \]
05

Conclusion

The series \( \sum_{n=1}^{\infty} \frac{7}{4^n} \) starts with terms \( \frac{7}{4}, \frac{7}{16}, \frac{7}{64}, \ldots \) and its sum is \( \frac{7}{3} \).

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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 a sum of terms that continues without end. Think of it as adding numbers forever, but interestingly, even these endless sums can have a definite result. In mathematics, when we write an infinite series like \( \sum_{n=1}^{\infty} a_n \), we're talking about adding an infinite number of terms \( a_1 + a_2 + a_3 + \ldots \) The series is made specific by the formula for each term.
In our example, \( \sum_{n=1}^{\infty} \frac{7}{4^n} \), each term is generated using the formula \( \frac{7}{4^n} \). This means we're adding forever but each time dividing by a larger and larger power of 4.
Common Ratio
The 'common ratio' is a key component in geometric series, playing a significant role in the sequence of terms. It refers to the consistent factor that each term is multiplied by to yield the next term. We denote this factor as \( r \).
  • If \( r \) is closer to zero, terms get smaller rapidly making the series converge quickly.
  • If \( r \) is negative, the terms alternate between positive and negative.
In our series, \( \sum_{n=1}^{\infty} \frac{7}{4^n}\), the common ratio is \( \frac{1}{4} \). This tells us that each term is one-quarter of the term before it, thus steadily decreasing.
Geometric Progression
A geometric progression is essentially a sequence where each term after the first is determined by multiplying the previous one by a fixed, non-zero number termed as the common ratio. This criterion gives geometric progressions their characteristic multiplying pattern.
  • Starting with a first term \( a \), each subsequent term is \( a \cdot r, a \cdot r^2, a \cdot r^3, \ldots \).
  • To visualize, our series produces the terms: \( \frac{7}{4}, \frac{7}{16}, \frac{7}{64}, \ldots \)
The series \( \sum_{n=1}^{\infty} \frac{7}{4^n} \) is a classic example of a geometric progression. Here, you can see how each term is derived from multiplying the previous term by the constant \( \frac{1}{4} \).
Series Convergence
For an infinite geometric series to be useful, we often want it to converge, or approach a specific number even as it stretches into infinity. A series converges when the sum of its infinite terms levels out at a certain value.
The convergence of a geometric series depends on the absolute value of the common ratio. If \( |r| < 1 \), the series converges.
  • A converging geometric series has a finite sum \( S \) given by the formula \( S = \frac{a}{1-r} \), where \( a \) is the first term.
  • Our series is convergent because \( |\frac{1}{4}| < 1 \).
Plugging \( \frac{7}{4} \) as the first term and \( \frac{1}{4} \) as the common ratio into the formula, we calculate the sum as \( \frac{7}{3} \). This showcases how even infinite series can have meaningful sums when they converge.

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