This is a geometric series with first term a = -4 and common ratio r = -(1/3).
The formula for the sum of an infinite geometric series is:
S = a/(1-r)
Plugging in the values, we get:
S = (-4)/(1-(-(1/3)))
S = (-4)/((4/3))
S = -3
Therefore, the series converges to -3.
Does this series converge or diverge? If it converges what is the sum? \sum_(n=1)^(\infty ) -4(-(1)/(3))^(n-1)
3 answers
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To show that the series converges to -3, we need to find the sum of the series.
The series can be written as:
-4(-1/3)^(0) - 4(-1/3)^(1) - 4(-1/3)^(2) - 4(-1/3)^(3) - ...
The first term is -4 and the common ratio is -1/3.
We know that a geometric series converges if and only if the absolute value of the common ratio is less than 1, that is, |r| < 1. In this case, |r| = |-1/3| = 1/3 < 1, so the series converges.
The formula for the sum of a convergent geometric series is:
S = a/(1 - r)
where a is the first term and r is the common ratio.
Plugging in the values, we get:
S = -4/(1 - (-1/3))
S = -4/(4/3)
S = -4 * 3/4
S = -3
Therefore, the sum of the series is -3 and the series converges.
The series can be written as:
-4(-1/3)^(0) - 4(-1/3)^(1) - 4(-1/3)^(2) - 4(-1/3)^(3) - ...
The first term is -4 and the common ratio is -1/3.
We know that a geometric series converges if and only if the absolute value of the common ratio is less than 1, that is, |r| < 1. In this case, |r| = |-1/3| = 1/3 < 1, so the series converges.
The formula for the sum of a convergent geometric series is:
S = a/(1 - r)
where a is the first term and r is the common ratio.
Plugging in the values, we get:
S = -4/(1 - (-1/3))
S = -4/(4/3)
S = -4 * 3/4
S = -3
Therefore, the sum of the series is -3 and the series converges.
4 answers