Problem 61
Question
Determine whether each infinite geometric series converges or diverges. If it converges, find its sum. $$ \sum_{k=1}^{\infty} 5\left(\frac{1}{4}\right)^{k-1} $$
Step-by-Step Solution
Verified Answer
The series converges, and its sum is \( \frac{20}{3} \).
1Step 1 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 \[ \frac{1}{4} \]
2Step 2 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.
3Step 3 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 \( \frac{20}{3} \).
Key Concepts
convergence of seriessum of geometric seriescommon ratio in sequences
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.
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 \( \frac{20}{3} \).
\[ \text{Sum} = \frac{a}{1 - r} \]
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.
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.
Other exercises in this chapter
Problem 59
How many terms must be added in an arithmetic sequence whose first term is 11 and whose common difference is 3 to obtain a sum of \(1092 ?\)
View solution Problem 60
Express each sum using summation notation. \(1^{3}+2^{3}+3^{3}+\cdots+8^{3}\)
View solution Problem 61
Find the vertical asymptotes, if any, of the graph of $$ f(x)=\frac{3 x^{2}}{(x-3)(x+1)} $$
View solution Problem 61
Express each sum using summation notation. \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{13}{13+1}\)
View solution