What is the sum from n=5 to infinity of 1/(2^{2n)} ?

The sum from n=5 to infinity of 1/(2^{2n)} is converges

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sum from n=5 to infinity of 1/(2^{2n)}

n = 5 \sum \infty \frac{1}{2 ^{2 n}}

converges

Solution steps

n = 5 \sum \infty \frac{1}{2 ^{2 n}}

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

= n = 5 \sum \infty \frac{1}{4 ^{n}}

Apply Series Ratio Test:

converges

= converges

f (x) = 5 - 8 x^{2}

\frac{d y}{d x} = 7 x y

\frac{d}{d x} (e^{3 x + 2 y})

f (x) = (7 x - 8)^{2}

\frac{d}{d y} (y - x)

What is the sum from n=5 to infinity of 1/(2^{2n)} ?
The sum from n=5 to infinity of 1/(2^{2n)} is converges