What is the sum from n=0 to infinity of 1/(2^{2n)} ?

The sum from n=0 to infinity of 1/(2^{2n)} is 4/3

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sum from n=0 to infinity of 1/(2^{2n)}

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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}

\int \frac{1}{6} e^{6 x} d x

x \to 3 lim (x^{2} - 3 x - 10)

x \to 0 + lim (9 sin (x) ln (x))

\int_{0}^{\frac{π}{9}} 7 - 7 cos (3 x) d x

x \to 1 lim (- \frac{x}{x ^{2} - x})

What is the sum from n=0 to infinity of 1/(2^{2n)} ?
The sum from n=0 to infinity of 1/(2^{2n)} is 4/3