#### Answer

a) First Four Terms: $\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81}$
b) $\lim\limits_{n \to \infty} S_n = \frac{1}{2}$

#### Work Step by Step

$$\sum_{k=1}^{\infty} 3^{-k}$$
Part A)
$a_1 = 3^{-1} = \frac{1}{3}$
$a_2 = 3^{-2} = \frac{1}{9}$
$a_3 = 3^{-3} = \frac{1}{27}$
$a_4 = 3^{-4} = \frac{1}{81}$
Part B)
This is a geometric series with an initial term of $a_1 = \frac{1}{3}$ and factor of $r = \frac{1}{3}$, which satisfies $r < 1$. Since it's a geometric series, a limit does exist.
$S_\infty = \frac{a_1}{1 - r} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}$