Step 1: Determine if if the infinite geometric series is divergence or convergence
If the common ratio is greater than or equals 1, then it is divergence but if r is less than 1, it is convergence.
[tex]\begin{gathered} \text{divergence if }\lvert r\rvert\ge1 \\ \text{convergence if }\lvert r\rvert<1 \end{gathered}[/tex]Step 2: Find common ratio r
[tex]\begin{gathered} r=\frac{a_2}{a_1}=\frac{a_3}{a_2}_{} \\ r=-\frac{1}{27}\times\frac{81}{1}=-\frac{81}{27}=-\frac{4}{3} \end{gathered}[/tex][tex]\lvert r\rvert=\lvert-\frac{4}{3}\rvert=\frac{4}{3}[/tex]From the value of the absolute value of r gotten, it can be observed that r is greater than 1 or equals 1, hence it is divergence.
Hence, the series will have an infinitely large sum. The correct option is FALSE
