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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y = 2 e^{- 5 x}

\frac{\partial}{\partial x} (3 x^{3} y)

y = (x^{2} + x)^{x}

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

y^{^{'}} = 2 x y^{\frac{3}{4}}

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