[tex] \bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\end{cases}\\\\\\~~~~~~~~~~~~~~~~~~\sum\limits_{n=2}^{5}~-3(\stackrel{\stackrel{r}{\downarrow }}{4})^{n-1} [/tex]
now, as you can see there, the common ratio is 4.
is it divergent or convergent?
well, the tale-tell fellow is the common ratio, if the common ratio is a fraction, is convergent, in this case it isn't, is 4, so is divergent.
a geometric serie is convergent only if the common ratio is a fraction, namely
0 < | r | < 1, is a value between 0 and 1.
