What is the sum from n=0 to infinity of 2^{-2n} ?

The sum from n=0 to infinity of 2^{-2n} is 4/3

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sum from n=0 to infinity of 2^{-2n}

n = 0 \sum \infty 2^{- 2 n}

\frac{4}{3}

Solution steps

n = 0 \sum \infty 2^{- 2 n}

Simplify

2^{- 2 n} : \frac{1}{2 ^{2 n}}

= n = 0 \sum \infty \frac{1}{2 ^{2 n}}

2^{2 n} = 4^{n}

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

Apply Series Geometric Test:

\frac{4}{3}

= \frac{4}{3}

x \to 0 lim (4^{2})

\frac{d}{d x} (\frac{36}{x})

\frac{3 s + 4}{s ^{2} - 1}

\int 3 e^{- x} - 2 sin (\frac{π}{2} - x) d x

\frac{d}{d x} (3 x^{2} - 6 x - 9)

What is the sum from n=0 to infinity of 2^{-2n} ?
The sum from n=0 to infinity of 2^{-2n} is 4/3