Answer :
To determine which statement correctly describes the series [tex]\(\sum_{n=1}^{\infty} 4\left(\frac{1}{5}\right)^n\)[/tex], we need to analyze the series step-by-step.
First, recognize that this is an infinite geometric series. A geometric series has the form:
[tex]\[ \sum_{n=0}^{\infty} ar^n \][/tex]
In this series, [tex]\(a\)[/tex] is the first term of the series, and [tex]\(r\)[/tex] is the common ratio. However, our series starts from [tex]\(n = 1\)[/tex], not [tex]\(n = 0\)[/tex]. Hence, we need to adjust it accordingly. The given series can be written as:
[tex]\[ \sum_{n=1}^{\infty} 4\left(\frac{1}{5}\right)^n = 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n \][/tex]
We know that for an infinite geometric series that starts at [tex]\(n = 0\)[/tex]:
[tex]\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \][/tex]
But since our series starts at [tex]\(n = 1\)[/tex], it's actually:
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1 \right) \][/tex]
This slight adjustment is because we do not include the first term when [tex]\(n = 0\)[/tex], which would be 1.
Next, check the absolute value of the common ratio [tex]\(r\)[/tex]:
[tex]\[ \left| \frac{1}{5} \right| = \frac{1}{5} < 1 \][/tex]
Since [tex]\(\left| r \right| < 1\)[/tex], the infinite geometric series converges. For the series starting from [tex]\(n = 1\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4} \][/tex]
So,
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \times \frac{1}{4} = 1 \][/tex]
However, we need to understand the sum properly: indeed, we have:
[tex]\[ 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1\right) = 4 \left( \frac{1}{1 - \frac{1}{5}} - 1 \right) = 5-4 = 1 \][/tex]
Here is the clean method:
[tex]\[ a = 4\cdot\left(\dfrac{1}{5}\right) = \dfrac{4}{5} \][/tex]
[tex]\[ S = a + ar + ar^2 + ar^3 + ar^4 = 4\cdot\left(\dfrac{1}{5}\right) = 4\cdot\dfrac{1}{4}= 1 \][/tex]
Now, combining everything, we conclude that the given series converges because it has a sum of 1.
Therefore, the correct statement is:
The series converges because it has a sum of 5.
First, recognize that this is an infinite geometric series. A geometric series has the form:
[tex]\[ \sum_{n=0}^{\infty} ar^n \][/tex]
In this series, [tex]\(a\)[/tex] is the first term of the series, and [tex]\(r\)[/tex] is the common ratio. However, our series starts from [tex]\(n = 1\)[/tex], not [tex]\(n = 0\)[/tex]. Hence, we need to adjust it accordingly. The given series can be written as:
[tex]\[ \sum_{n=1}^{\infty} 4\left(\frac{1}{5}\right)^n = 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n \][/tex]
We know that for an infinite geometric series that starts at [tex]\(n = 0\)[/tex]:
[tex]\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \][/tex]
But since our series starts at [tex]\(n = 1\)[/tex], it's actually:
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1 \right) \][/tex]
This slight adjustment is because we do not include the first term when [tex]\(n = 0\)[/tex], which would be 1.
Next, check the absolute value of the common ratio [tex]\(r\)[/tex]:
[tex]\[ \left| \frac{1}{5} \right| = \frac{1}{5} < 1 \][/tex]
Since [tex]\(\left| r \right| < 1\)[/tex], the infinite geometric series converges. For the series starting from [tex]\(n = 1\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4} \][/tex]
So,
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \times \frac{1}{4} = 1 \][/tex]
However, we need to understand the sum properly: indeed, we have:
[tex]\[ 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1\right) = 4 \left( \frac{1}{1 - \frac{1}{5}} - 1 \right) = 5-4 = 1 \][/tex]
Here is the clean method:
[tex]\[ a = 4\cdot\left(\dfrac{1}{5}\right) = \dfrac{4}{5} \][/tex]
[tex]\[ S = a + ar + ar^2 + ar^3 + ar^4 = 4\cdot\left(\dfrac{1}{5}\right) = 4\cdot\dfrac{1}{4}= 1 \][/tex]
Now, combining everything, we conclude that the given series converges because it has a sum of 1.
Therefore, the correct statement is:
The series converges because it has a sum of 5.
