Does this series converge or diverge? If it converges what is the sum? \sum_(n=1)^(\infty ) -4(-(1)/(3))^(n-1)
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.
show your work
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.
To determine whether the series \(\sum_{n=1}^{\infty} -4\left(-\frac{1}{3}\right)^{n-1}\) converges or diverges, we can use the formula for the sum of an infinite geometric series.
For a geometric series in the form \(\sum_{n=1}^{\infty} ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio, the series converges if and only if \(\left| r \right| < 1\). In that case, the sum can be found using the formula:
S = a / (1 - r)
In this case, \(a = -4\) and \(r = -\frac{1}{3}\). Let's substitute these values into the formula for the sum:
S = -4 / (1 - \left(-\frac{1}{3}\right))
= -4 / \left(1 + \frac{1}{3}\right)
= -4 / \left(\frac{4}{3}\right)
= - 4 \cdot \frac{3}{4}
= -3
Therefore, the series converges, and the sum of the series is -3.