Notice that in both cases we have geometric series. Then, we have the following general rule:
[tex]\begin{gathered} \Sigma_{k=0}^{\infty}(ar^k)=\frac{a}{1-r} \\ if\text{ \mid r\mid<1} \end{gathered}[/tex]
then, in the first case we have the following sum:
[tex]\Sigma_{k=0}^{\infty}3(-\frac{1}{5})^k=\frac{3}{1-(-\frac{1}{5})}=\frac{3}{1+\frac{1}{5}}=\frac{3}{\frac{6}{5}}=\frac{15}{6}=\frac{5}{2}[/tex]
for the second case, we hve:
[tex]\Sigma_{k=1}^{\infty}2(-\frac{3}{5})^k=\frac{2(-\frac{3}{5})}{1-(-\frac{3}{5})}=\frac{-\frac{6}{5}}{1+\frac{3}{5}}=\frac{-\frac{6}{5}}{\frac{8}{5}}=\frac{-30}{40}=-\frac{3}{4}[/tex]
