What is the sum from n=0 to infinity of 4^{-n} ?

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

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sum from n=0 to infinity of 4^{-n}

n = 0 \sum \infty 4^{- n}

\frac{4}{3}

Solution steps

n = 0 \sum \infty 4^{- n}

Simplify

4^{- n} : \frac{1}{4 ^{n}}

= n = 0 \sum \infty \frac{1}{4 ^{n}}

Apply Series Geometric Test:

\frac{4}{3}

= \frac{4}{3}

\frac{d}{d y} (y cos (x) + 2 x e^{y})

2 y^{^{''}} - 4 y^{^{'}} + 8 y = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = t

f (x) = 3 (x)^{2} - 2 (x)

\frac{d y}{d x} = 4 - \frac{5 y}{300 - 3 x}

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