What is the sum from n=0 to infinity of (-3)^{-n} ?

The sum from n=0 to infinity of (-3)^{-n} is 3/4

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sum from n=0 to infinity of (-3)^{-n}

n = 0 \sum \infty (- 3)^{- n}

\frac{3}{4}

Decimal

0.75

Solution steps

n = 0 \sum \infty (- 3)^{- n}

Simplify

(- 3)^{- n} : \frac{1}{( - 3 ) ^{n}}

= n = 0 \sum \infty \frac{1}{( - 3 ) ^{n}}

Apply Series Geometric Test:

\frac{3}{4}

= \frac{3}{4}

What is the sum from n=0 to infinity of (-3)^{-n} ?
The sum from n=0 to infinity of (-3)^{-n} is 3/4