Question

Determine whether the infinite geometric series:

$256-64+16-4+...$

converges or diverges. If it converges, find its sum.

Short answer

$256-64+16-4+...$converges.

$S_{\infty }=\frac{1024}{5}$

Step-by-step solution

Given information:

Consider the series,$256-64+16-4+...$

Formulas:

$$\begin{gathered} \left(1\right)Commonratio:r=\frac{a_n}{a_{n-1}} \\ \left(2\right)S_{\infty }=\frac{a_1}{1-r} \end{gathered}$$

To check whether the series converges or diverges:

Given Series, $256-64+16-4+...$

Rewriting the sum of geometric series:$256+(-64)+16+(-4)+...$

Here, $a_1=256,a_2=-64,a_3=16,a_4=-4,...$

$$\begin{gathered} r=\frac{-64}{256}=\frac{-1}{4} \\ r=\frac{16}{-64}=\frac{-1}{4} \\ r=\frac{-4}{16}=\frac{-1}{4} \end{gathered}$$

Common Ratio, $r=\frac{-1}{4}=-0.75<1$

Since $\left|r\right|<1$, the series converges.

To find its sum:

Here, $a_1=256,r=\frac{-1}{4}$

$$\begin{gathered} S_{\infty }=\frac{256}{1-\left({\displaystyle \frac{-1}{4}}\right)}\left[Useformula\left(2\right)\right] \\ =\frac{256}{1+{\displaystyle \frac{1}{4}}} \\ =\frac{256}{{\displaystyle \frac{4+1}{4}}} \\ =256\times \frac{4}{5} \\ =\frac{1024}{5} \end{gathered}$$