Question

consider the infinite geometric series.in this image the lower limit of the summation 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.

Answer 1

The geometric series is;

`s_n=\sum ^(\infty)_(n\mathop=1)-4((1)/(3))^(n-1)`

when n=1, we have

`s_1=-4((1)/(3))^(1-1)=-4((1)/(3))^0=-4*1=-4`

When n=2, we have,

`s_2=-4((1)/(3))^(2-1)=-4((1)/(3))^1=-4*(1)/(3)=-(4)/(3)`

When n = 3, we get,

`s_3=-4((1)/(3))^(3-1)=-4((1)/(3))^2=-4*(1)/(9)=-(4)/(9)`

when n = 4, we get,

`s_4=-4((1)/(3))^(4-1)=-4*((1)/(3))^3=-(4)/(27)`

a. So, the first four terms of the series are: - 4, -4/3, -4/9 and -4/27

The sum to infinity of the series is:

`S_(\infty)=(a)/(1-r)=(-4)/(1-(1)/(3))=(-4)/((2)/(3))=(-4*3)/(2)=-2*3=-6`

b. The series, as we can see, is CONVERGENT, because it has a definite value as its sum.

c. The sum of the series is - 6