Does this series converge or diverge? If it converges what is the sum? \sum_(n=1)^(\infty ) -4(-(1)/(3))^(n-1)

3 answers

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.