# Find the infinite sum, if it exists: –2+1/2-1/8+1/32-...?

Apr 16, 2017

${\text{S}}_{\infty} = - \frac{8}{5}$

#### Explanation:

For a geometric sequence, $a , a r , a {r}^{2.} . .$ where $a$ is the first term and $r$ is the common ratio between two consecutive terms, the sum to infinity, ${\text{S}}_{\infty} ,$ is given in this equation, ${\text{S}}_{\infty} = \frac{a}{1 - r}$.

The given series is geometric because the common ratio, $r$, is $- \frac{1}{4}$.

Therefore, ${\text{S}}_{\infty} = \frac{- 2}{1 - - \frac{1}{4}} = - \frac{8}{5}$

Apr 16, 2017

The infinite sum exists; S=–8/5.

#### Explanation:

The sequence converges, since each term is $- \frac{1}{4}$ times the previous term, and $\left\mid - \frac{1}{4} \right\mid < 1$. Thus the infinite sum exists. Let us call this sum $S$. Then:

color(white)(–1/2)S=-2+1/2-1/8+...

color(white)(–1/2S)=-2(1-1/4+1/16-1/64+...)

color(blue)(–1/2 S="    "1-1/4+1/16-1/64+...)

If we multiply both sides by $- \frac{1}{4}$, we get

(–1/4)(–1/2 S)=(–1/4)(1-1/4+1/16-1/64+...)

$\textcolor{g r e e n}{\text{ } \frac{1}{8} S = - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \ldots}$

Notice how the expansion for $\textcolor{b l u e}{- \frac{1}{2} S}$ includes the expansion for $\textcolor{g r e e n}{\frac{1}{8} S}$. If we subtract $\frac{1}{8} S$ from $- \frac{1}{2} S$ we get:

"      "color(blue)(–1/2 S) - color(green)(1/8 S)=color(blue)("   "1-1/4+1/16-1/64+...)
color(white)(–1/2 S - 1/8 S=)color(green)("     "-(-1/4+1/16-1/64+...))

"                  "–5/8 S=1

Solving for $S$ gives:

$S = - \frac{8}{5}$