Which of the following geometric series converges?

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
[tex]S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}[/tex]
Multiply both sides by r :
[tex]rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n[/tex]
Subtract the latter sum from the first, which eliminates all but the first and last terms:
[tex]S_n-rS_n=a-ar^n[/tex]
Solve for [tex]S_n[/tex]:
[tex](1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}[/tex]
Then as gets arbitrarily large, the term [tex]r^n[/tex] will converge to 0, leaving us with
[tex]S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}[/tex]
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