# How do you find the fourth partial sum of Sigma 8(-1/4)^n from n=1 to oo?

Jul 7, 2017

$- \frac{51}{32}$

#### Explanation:

Let use denote ${S}_{n}$ by:

${S}_{n} = {\sum}_{r = 1}^{n} \setminus 8 {\left(- \frac{1}{4}\right)}^{r}$

So then. the fourth partial sum, is ${S}_{4}$:

${S}_{4} = {\sum}_{r = 1}^{4} \setminus 8 {\left(- \frac{1}{4}\right)}^{r}$
$\text{ } = 8 {\left(- \frac{1}{4}\right)}^{1} + 8 {\left(- \frac{1}{4}\right)}^{2} + 8 {\left(- \frac{1}{4}\right)}^{3} + 8 {\left(- \frac{1}{4}\right)}^{4}$
$\text{ } = 8 \left(- \frac{1}{4}\right) + 8 \left(\frac{1}{16}\right) + 8 \left(- \frac{1}{64}\right) + 8 \left(\frac{1}{256}\right)$
$\text{ } = 8 \left\{- \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \frac{1}{256}\right)$
$\text{ } = 8 \left\{- \frac{51}{256}\right)$
$\text{ } = - \frac{51}{32}$