Question

Which of the following geometric series converges?

Answer 1

All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term a and common ratio |r| < 1. Then the n-th partial sum (the sum of the first n terms) of the sequence is

`S_n=a+ar+ar^2+\cdots+ar^(n-2)+ar^(n-1)`

Multiply both sides by r :

`rS_n=ar+ar^2+ar^3+\cdots+ar^(n-1)+ar^n`

Subtract the latter sum from the first, which eliminates all but the first and last terms:

`S_n-rS_n=a-ar^n`

Solve for
`S_n`:

`(1-r)S_n=a(1-r^n)\implies S_n=\frac a{1-r}-(ar^n)/(1-r)`

Then as gets arbitrarily large, the term
`r^n` will converge to 0, leaving us with

`S=\displaystyle\lim_(n\to\infty)S_n=\frac a{1-r}`

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18