How do you find the sum of the infinite geometric series 1-(1/3) + (1/9) - (1/27) + .....?

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Answer 1 · 2016-02-22T01:01:25

The formula for sum of an infinite geometric series is $s_{\infty} = \frac{a}{1 - r}$ $s_oo = a/(1 - r)$

We must first find the common ratio, r, which can be found by using the formula $r = \frac{t_{2}}{t_{1}}$ $r = t_2/t_1$

$r = \frac{- \frac{1}{3}}{1}$ $r = (-1/3)/1$

$r = - \frac{1}{3}$ $r = -1/3$

a is the first term, 1.

$s_{\infty} = \frac{1}{1 - (- \frac{1}{3})}$ $s_oo = 1/(1 - (-1/3))$

$s_{\infty} = \frac{1}{\frac{4}{3}}$ $s_oo = 1/(4/3)$

$s_{\infty} = \frac{3}{4}$ $s_oo = 3/4$

The sum is $\frac{3}{4}$ $3/4$ .

Hopefully this helps!