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

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

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sum from n=0 to infinity of 1/(4^{n+1)}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (e^{x^{2} - 2 x + 1})

\frac{\partial}{\partial x} (x^{3} y^{2} e^{x z})

\frac{d}{d x} (\frac{1 + e ^{- 2 x}}{2 + cos ( 2 x )})

\int (cos (\frac{x}{2}))^{2} d x

f (x) = 3 x^{3} - x^{2} + 1, (2, 21)

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