Consider the infinite geometric series
∑^(∞)_(n=1) −4(1/3)^n−1
. In this image, the lower limit of the summation notation is "n = 1".
a. Write the first four terms of the series.
b. Does the series diverge or converge?
c. If the series has a sum, find the sum.
The series is
∑^(∞)_(n=1) −4(1/3)^n−1 = -4 -4/3 -4/3^2 - 4/3^3 - 4/3^4
(first four terms are listed)
b. The series converges because this is a geometric series.
4= 1/3 <1
the sum is -6
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Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1
a. Write the first four terms of the series.
b. Does the series diverge or converge.
c. If the series has a sum, find the sum.
a. Sure, here are the first four terms of the series:
-4(1/3)^0, -4(1/3)^1, -4(1/3)^2, -4(1/3)^3
b. Well, let's take a closer look. The common ratio of this series is (1/3). When the absolute value of the common ratio is less than 1, the series converges. In this case, the absolute value of the common ratio is 1/3, which is indeed less than 1. So, this series converges!
c. To find the sum of this infinite geometric series, we can use the formula for the sum of a convergent geometric series:
Sum = a / (1 - r)
where "a" is the first term and "r" is the common ratio. Plugging in the values from our series, we get:
Sum = -4 / (1 - 1/3)
Simplifying further, we have:
Sum = -4 / (2/3)
Sum = -6
So, the sum of this series is -6. Just be careful not to drop it!
a. To write the first four terms of the series, we substitute n = 1, 2, 3, and 4 into the given formula.
When n = 1: -4(1/3)^(1-1) = -4(1/3)^0 = -4(1) = -4
When n = 2: -4(1/3)^(2-1) = -4(1/3)^1 = -4(1/3) = -4/3
When n = 3: -4(1/3)^(3-1) = -4(1/3)^2 = -4(1/9) = -4/9
When n = 4: -4(1/3)^(4-1) = -4(1/3)^3 = -4(1/27) = -4/27
Therefore, the first four terms of the series are -4, -4/3, -4/9, and -4/27.
b. To determine if the series converges or diverges, we need to calculate the common ratio, |r|, which is the absolute value of the ratio between consecutive terms. In this case, the common ratio is 1/3.
Since |r| < 1, the series converges.
c. To find the sum of the series, we use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where S is the sum of the series, a is the first term, and r is the common ratio.
In this case, a = -4 and r = 1/3. Substituting the values into the formula, we have:
S = -4 / (1 - 1/3)
S = -4 / (2/3)
S = -4 * (3/2)
S = -12/2
S = -6
Therefore, the sum of the series is -6.
To answer these questions about the infinite geometric series, let's first understand the general form of an infinite geometric series:
∑^(∞)_(n=1) ar^(n-1)
In this formula, "a" represents the first term of the series and "r" represents the common ratio between consecutive terms.
a. To write the first four terms of the given series, we substitute the values:
a = -4
r = 1/3
So, the first term (n = 1) is -4(1/3)^(1-1) = -4. The second term (n = 2) is -4(1/3)^(2-1) = -4(1/3) = -4/3. The third term (n = 3) is -4(1/3)^(3-1) = -4(1/9) = -4/9. The fourth term (n = 4) is -4(1/3)^(4-1) = -4(1/27) = -4/27.
Hence, the first four terms of the series are: -4, -4/3, -4/9, -4/27.
b. To determine whether the series converges or diverges, we need to evaluate the common ratio "r". If the absolute value of "r" is less than 1, the series converges. Otherwise, if the absolute value of "r" is greater than or equal to 1, the series diverges.
In this case, |1/3| < 1, so the series converges.
c. To find the sum of the series, we can use the formula for the sum of an infinite geometric series:
S = a/(1 - r)
Plugging in the given values:
a = -4
r = 1/3
Substituting the values, we have:
S = -4 / (1 - 1/3)
= -4 / (2/3)
= -4 * (3/2)
= -12/2
= -6
Hence, the sum of the given series is -6.
(a) really? Just plug in the values for n
-4/3^0, -4/3^1, -4/3^2, -4/3^3
(b) this is just a GP with r = 1/3
(c) S = a/(1-r) = -4/(1 - 1/3) = -6