How do you find the sum of the infinite geometric series 1/8-1/16+1/32-1/64+...?

Question

9023 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-28T08:21:00

$S_{\infty} = \frac{1}{12}$ $S_oo=1/12$

#1/8-1/16+1/32-1/64--(1)

a GP is in the form

$a + a r + a r^{2} + . . a r^{n - 1} + . .$ $a+ar+ar^2+..ar^(n-1)+..$

for $(1)$ $(1)$

we have $a = \frac{1}{8}$ $a=1/8$

to find the common ratio $r$ $r$ we divide adjacent terms

$r = \frac{- \frac{1}{16}}{\frac{1}{8}} = - \frac{1}{2}$ $r=(-1/16)/(1/8)=-1/2$

for a GP to have a sum to infinity

$| r | < 1$ $|r|<1$

$| r | = ∣ ∣ ∣ - \frac{1}{2} ∣ ∣ ∣ < 1$ $|r|=|-1/2|<1$

$:.EES_oo$

$S_oo=a/(1-r)$

$S_oo=(1/8)/(1- -1/2)$

$S_oo=(1/8)/((3/2))=1/8-:3/2$

$S_oo=1/12$