Question

Consider the infinite geometric series (∞/Σ/n=1) * -4(1/3)^{n-1}.

Answer 1

The first four terms of the series probably refer to the first four partial sums.

$\displaystyle\sum_{n=1}^1-4\left(\frac13\right)^{n-1}=-4\left(\frac13\right)^{1-1}=-4$
$\displaystyle\sum_{n=1}^2-4\left(\frac13\right)^{n-1}=-\frac{16}3$
$\displaystyle\sum_{n=1}^3-4\left(\frac13\right)^{n-1}=-\frac{52}9$
$\displaystyle\sum_{n=1}^4-4\left(\frac13\right)^{n-1}=-\frac{160}{27}$

which you can compute either by adding one term at a time, or using the well-known formula,

$\displaystyle\sum_{n=1}^kar^{n-1}=a\frac{1-r^{k+1}}{1-r}$

(I can provide a link to a derivation I gave in a nearly identical question in the comments)

The series converges as $k\to\infty$ if and only if $|r|$, which is certainly the case here, since $-1$.

Extrapolating from the formula above, the sum of the convergent series is

$\displaystyle\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}$

so the sum for this series is $\dfrac{-4}{1-\frac13}=-6$.