# How to show that S_oo- S_n= (-1/3)^n ,given, S_oo=9/4 and S_n=9/4(1-(-1/3)^n)?

Jun 21, 2018

I think there's a $\frac{9}{4}$ factor missing

#### Explanation:

Since you're given both ${S}_{\setminus} \infty$ and ${S}_{n}$, in order to compute ${S}_{\setminus} \infty - {S}_{n}$ you can simply subtract the two expressions:

color(red)(S_\infty)-color(blue)(S_n) = color(red)(9/4) - color(blue)(9/4(1-(-1/3)^n)

We can expand ${S}_{n}$ as follows:

$\frac{9}{4} \left(1 - {\left(- \frac{1}{3}\right)}^{n}\right) = \frac{9}{4} - \frac{9}{4} \setminus \times {\left(- \frac{1}{3}\right)}^{n}$

So, we have

${S}_{\setminus} \infty - {S}_{n}$
$= \frac{9}{4} - \left(\frac{9}{4} - \frac{9}{4} \setminus \times {\left(- \frac{1}{3}\right)}^{n}\right)$
$= \cancel{\frac{9}{4}} - \cancel{\frac{9}{4}} + \frac{9}{4} \setminus \times {\left(- \frac{1}{3}\right)}^{n}$

So,

${S}_{\setminus} \infty - {S}_{n} = \frac{9}{4} \setminus \times {\left(- \frac{1}{3}\right)}^{n}$