Consider the following geometric series. [infinity]∑n=1 (−8)n−19n
a) Find the common ratio.
b) Determine whether the geometric series is convergent or divergent.
c) If it is convergent, find its sum. (If the quantity diverges, answer diverges.)

a) -8/9

b) The series is a convergent series

c) 1/17

Step-by-step explanation:

The series a+ar+ar²+ar³⋯ =∑ar^(n−1) is called a geometric series, and r is called the common ratio.

If −1<r<1, the geometric series is convergent and its sum is expressed as ∑ar^(n−1) = a/1-r

a is the first tern of the series.

a) Rewriting the series ∑(-8)^(n−1)/9^n given in the form ∑ar^(n−1) we have;

∑(-8)^(n−1)/9^n

= ∑(-8)^(n−1)/9•(9)^n-1

= ∑1/9 • (-8/9)^(n−1)

From the series gotten, it can be seen in comparison that a = 1/9 and r = -8/9

The common ratio r = -8/9

b) Remember that for the series to be convergent, -1<r<1 i.e r must be less than 1 and since our common ratio which is -8/9 is less than 1, this implies that the series is convergent.

c) Since the sun of the series tends to infinity, we will use the formula for finding the sum to infinity of a geometric series.

S∞ = a/1-r

Given a = 1/9 and r = -8/9

S∞ = (1/9)/1-(-8/9)

S∞ = (1/9)/1+8/9

S∞ = (1/9)/17/9

S∞ = 1/9×9/17

S∞ = 1/17

The sum of the geometric series is 1/17

3√128

Step-by-step explanation: