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Chapter 13: Problem 61

Determine whether each infinite geometric series converges or diverges. If it converges, find its sum. $$ \sum_{k=1}^{\infty} 5\left(\frac{1}{4}\right)^{k-1} $$

Short Answer

Expert verified
The series converges, and its sum is \( \frac{20}{3} \).

Step by step solution

01

Title - Identify the Initial Term and Common Ratio

For an infinite geometric series in the form \ \[ \text{Sum} = \text{a}r^{(k-1)} \] where 'a' is the initial term and 'r' is the common ratio. Here, the series is \[ \text{Sum} = 5\bigg(\frac{1}{4}\bigg)^{k-1} \] Thus, the initial term 'a' is 5 and the common ratio 'r' is \[ \frac{1}{4} \]
02

Title - Check if the Series Converges or Diverges

To determine if the series converges or diverges, examine the common ratio 'r'. If \( |r| < 1 \) , then the geometric series converges. Otherwise, it diverges. In this case, \( |r| = \bigg|\frac{1}{4}\bigg| = 0.25 < 1 \). Therefore, the series converges.
03

Title - Calculate the Sum of the Converging Series

If a geometric series converges, its sum can be found using the formula \[ \text{Sum} = \frac{a}{1 - r} \] . Plugging in the values for 'a' and 'r', we get \[ \text{Sum} = \frac{5}{1 - \frac{1}{4}} = \frac{5}{\frac{3}{4}} = 5 \times \frac{4}{3} = \frac{20}{3} \]. Therefore, the sum of the series is \( \frac{20}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

convergence of series
For an infinite geometric series to either converge or diverge, we examine the common ratio, denoted as 'r'. If the absolute value of 'r' is less than 1, \( |r| < 1 \), then the series converges. This means the infinite sum approaches a finite value as more terms are added.

Conversely, if the absolute value of the common ratio is greater than or equal to 1, \( |r| \geq 1 \) , the series diverges. This implies the sum does not approach a finite limit. In our example, the common ratio is \( \frac{1}{4} \), with an absolute value of 0.25. Since 0.25 is less than 1, the series converges.
sum of geometric series
Once we've established that our geometric series converges, we can calculate its sum using a specific formula. The sum of an infinite geometric series, where \( |r| < 1 \), is given by:

\[ \text{Sum} = \frac{a}{1 - r} \]
Here, 'a' represents the initial term, and 'r' represents the common ratio. This formula ensures we can find a finite sum for an otherwise infinite series.

In our example, the initial term 'a' is 5, and the common ratio 'r' is \( \frac{1}{4} \). Plugging these values into the formula, we get:

\[ \text{Sum} = \frac{5}{1 - \frac{1}{4}} = \frac{5}{\frac{3}{4}} = 5 \times \frac{4}{3} = \frac{20}{3} \]
So, the sum of this series is \( \frac{20}{3} \).
common ratio in sequences
The common ratio in a geometric sequence is a crucial element as it determines the behavior of the series. It is found by dividing any term in the series by the preceding term. Specifically, in our example, it can be calculated as follows:

Given the series: \( 5, \frac{5}{4}, \frac{5}{16}, ... \), the common ratio 'r' is:

\[ r = \frac{\frac{5}{4}}{5} = \frac{5 / 4}{5 / 1} = \frac{1}{4}\]

Understanding the common ratio helps us to determine if the series converges or diverges. If \( |r| < 1 \) it converges, and if \( |r| \geq 1 \), it diverges. In this case, \( \frac{1}{4} \) is our common ratio, confirming the convergence of the series.

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Most popular questions from this chapter

Bode's Law In \(1772,\) Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2} $$ where \(n \geq 2\) is the number of the planet from the sun. (a) Determine the first eight terms of the sequence. (b) At the time of Bode's publication, the known planets were Mercury \((0.39 \mathrm{AU}),\) Venus \((0.72 \mathrm{AU}),\) Earth \((1 \mathrm{AU})\) Mars \((1.52 \mathrm{AU}),\) Jupiter \((5.20 \mathrm{AU}),\) and Saturn \((9.54 \mathrm{AU})\) How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in \(1781,\) and the asteroid Ceres was discovered in \(1801 .\) The mean orbital distances from the sun to Uranus and Ceres " are \(19.2 \mathrm{AU}\) and \(2.77 \mathrm{AU},\) respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (e) The planets Neptune and Pluto" were discovered in 1846 and \(1930,\) respectively. Their mean orbital distances from the sun are \(30.07 \mathrm{AU}\) and \(39.44 \mathrm{AU},\) respectively. How do these actual distances compare to the terms of the sequence? (f) On July \(29,2005,\) NASA announced the discovery of a dwarf planet \((n=11),\) which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

Expand each expression using the Binomial Theorem. $$ \left(x^{2}-y^{2}\right)^{6} $$

Reflections in a Mirror A highly reflective mirror reflects \(95 \%\) of the light that falls on it. In a light box having walls made of the mirror, the light reflects back-and-forth between the mirrors. (a) If the original intensity of the light is \(I_{0}\) before it falls on a mirror, write the \(n\) th term of the sequence that describes the intensity of the light after \(n\) reflections. (b) How many reflections are needed to reduce the light intensity by at least \(98 \% ?\)

Find the fifth term and the nth term of the geometric sequence whose first term \(a_{1}\) and common ratio \(r\) are given. $$ a_{1}=2 ; \quad r=3 $$

Problems \(75-84\) are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. If a credit card charges \(15.3 \%\) interest compounded monthly, find the effective rate of interest.
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