By the given series have:
a_{n}=\frac{(-3)^{n-1}}{4^{n}}$, $a_{n+1}=\frac{(-3)^{n}}{4^{n+1}}
By the criterion of Dalamber have:
A=\lim\frac{a_{n+1}}{a_{n}}=\frac{3}{4}<1
Under this criterion we have that A<1 conclude that given series is convergent.
Since the given series is convergent exist sum of this series. Mark wis S_{n}$ sum of this series, it is $S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}.
Hance we
S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}
S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty \frac{(-3)^{n}}{4^n}
\frac{(-3)^{n}}{4^n} is it a geometric series. Now find sum this series. Find the sum must assign a_{1} dhe q.
a_{1}=-\frac{3}{4}$, $q=-\frac{3}{4}.
Sum accounst
S_{n}=\frac {a_{1}(1-q^{n})}{1-q}=-\frac{3}{7}{[1-\frac{3^{n}}{4^{n}}]}
\lim S_{n}=-\frac{3}{7}
Theres definitely have
S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty\frac{(-3)^{n}}{4^{n}}
S_{n}=(-\frac{1}{3})(-\frac{3}{7})
S_{n}=\frac{1}{7}
Conlude the: Given series is the convergent and its sum \frac{1}{7}$.$