Determine whether $\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$ is convergent or divergent. If convergent, find the sum. $$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$
It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = \frac{(-3)^{x-1}}{4^x}$.
$$\lim\limits_{x \to \infty}\frac{(-3)^{x-1}}{4^x}$$
Is this how I handle this exercise? I still cannot seem to get the answer $\frac{1}{7}$
 A: A geometric series is convergent if the $|r|<1$ where $r$ is the common ratio.
Let $S_n=\sum_{i=0}^n (-3/4)^i$ then $$S_n=\frac{(-3/4)^{n+1}-1}{(-3/4)-1}$$
Now take $n\rightarrow \infty$ then $$S_n\rightarrow \frac{0-1}{(-3/4)-1}=4/7$$
because $|-3/4|<1$ and so $(-3/4)^n\rightarrow 0$.
Now note that your sum is $$\mbox{lim }\sum_{i=1}^{n+1}\frac{(-3)^{i-1}}{4^{i}}=\mbox{lim }\frac{1}{4}\sum_{i=1}^{n+1}\frac{(-3)^{i-1}}{4^{i-1}}=1/4.\mbox{lim }S_n=1/7$$
A: If $\,a, ar, ar^2,...\,$ is a geometric series with $\,|r|<1\,$ ,then
$$\sum_{n=0}^\infty ar^n=\lim_{n\to\infty} ar^n=\lim_{n=0}\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}$$since $\,r^n\xrightarrow [n\to\infty]{} 0\Longleftrightarrow |r|<1\,$ , and thus
$$\sum_{n=1}^\infty\frac{(-3)^{n-1}}{4^n}=\frac{1}{4}\sum_{n=0}^\infty \left(-\frac{3}{4}\right)^n=\frac{1}{4}\frac{1}{1-\left(-\left(\frac{3}{4}\right)\right)}=\frac{1}{4}\frac{1}{\frac{7}{4}}=\frac{1}{7}$$
A: Let $q = \frac{-3}{4}$, $a_n = \frac{(-3)^{n-1}}{4^n}$, $b_0 = 0$, $b_n = b_{n-1} + a_n$.
$a_n = -\frac{1}{3} (\frac{-3}{4})^n = -\frac{1}{3} q^n$, hence $b_n = -\frac{1}{3} c_n$, where $c_0 = 0$, $c_n = c_{n-1} + q^n$.
The $c_n$ limit is equal to $q + q^2 + q^3 + \ldots = \frac{q}{1-q} = \frac{\frac{-3}{4}}{1-\frac{-3}{4}} = \frac{-3}{7}$, thus $b_n$ limit is equal to $\frac{1}{7}$.
A: $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 $\frac{1}{7}$.$
