Chapter 9: Problem 4
Determine whether the series converges, and if so find its sum. $$ \sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^{k+2} $$
Short Answer
Expert verified
The series converges and its sum is \( \frac{4}{3} \).
Step by step solution
01
Identify the Series Type
The series given is \( \sum_{k=1}^{\infty} \left(\frac{2}{3}\right)^{k+2} \). This is a geometric series because each term can be written as \( ar^{k} \) where \( a \) is the first term and \( r \) is the common ratio.
02
Express the Series in Standard Form
We can rewrite the general term \( \left(\frac{2}{3}\right)^{k+2} \) as \( \left(\frac{2}{3}\right)^2 \left(\frac{2}{3}\right)^{k} \). This means the series can be written as \( \sum_{k=1}^{\infty} \left(\frac{4}{9}\right) \left(\frac{2}{3}\right)^k \), identifying \( a = \left(\frac{4}{9}\right) \) and \( r = \left(\frac{2}{3}\right) \).
03
Determine Convergence of the Series
For an infinite geometric series \( \sum_{k=0}^{\infty} ar^k \), it converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( |\frac{2}{3}| = \frac{2}{3} < 1 \), hence the series converges.
04
Calculate the Sum of the Series
The sum of a converging infinite geometric series is given by \( \frac{a}{1-r} \). In our case, \( a = \frac{4}{9} \) and \( r = \frac{2}{3} \). Thus, the sum is: \[ \frac{\frac{4}{9}}{1-\frac{2}{3}} = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3}. \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
series convergence
In the world of series, particularly geometric series, convergence is a critical concept to understand. A series is said to converge if it approaches a specific value as we sum progressively more terms. For a geometric series, which takes the form \( \sum_{k=0}^{\infty} ar^k \), convergence depends on the common ratio, \( r \).
The key condition for the convergence of an infinite geometric series is that the absolute value of the common ratio must be less than one:
The key condition for the convergence of an infinite geometric series is that the absolute value of the common ratio must be less than one:
- If \( |r| < 1 \), the series converges.
- If \( |r| \geq 1 \), the series diverges.
sum of infinite series
Once we've determined that a geometric series converges, we can calculate its sum. The sum of a convergent infinite geometric series \( \sum_{k=0}^{\infty} ar^k \) is calculated using the formula: \[ \text{Sum} = \frac{a}{1-r} \] where \( a \) is the first term of the series, and \( r \) is the common ratio.
In the exercise, we identified \( a = \frac{4}{9} \) and \( r = \frac{2}{3} \), leading us to the sum: \[ \frac{\frac{4}{9}}{1-\frac{2}{3}} = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3} \] The beauty of this is that we can find the sum of infinitely many terms in a series by using this simple formula, provided the series converges.
In the exercise, we identified \( a = \frac{4}{9} \) and \( r = \frac{2}{3} \), leading us to the sum: \[ \frac{\frac{4}{9}}{1-\frac{2}{3}} = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3} \] The beauty of this is that we can find the sum of infinitely many terms in a series by using this simple formula, provided the series converges.
common ratio
The concept of a common ratio is essential in understanding geometric series. The common ratio, denoted \( r \), is the factor by which we multiply each term to get the next one in the series. This ratio remains constant throughout the series.
To identify \( r \), we look at the expression for the series: each term is \( ar^k \). For our given series, we rewrote the general term \( \left(\frac{2}{3}\right)^{k+2} \) to reveal the form \( \left(\frac{4}{9}\right)\left(\frac{2}{3}\right)^k \), showing that \( r = \frac{2}{3} \).
To identify \( r \), we look at the expression for the series: each term is \( ar^k \). For our given series, we rewrote the general term \( \left(\frac{2}{3}\right)^{k+2} \) to reveal the form \( \left(\frac{4}{9}\right)\left(\frac{2}{3}\right)^k \), showing that \( r = \frac{2}{3} \).
- The common ratio determines whether the series converges.
- The magnitude of \( r \) affects the speed at which the series sum settles into its final value if it converges.