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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

e^{- s} \frac{1}{s - 4}

x \to \infty + lim (\frac{2 x + 1}{4 x ^{2} + x})

\frac{y}{x} \cdot \frac{d y}{d x} = - 4 - \frac{y ^{2}}{x ^{2}}

x \to 3 lim (\frac{1}{x})

x y^{^{'}} - y = \frac{x ^{2}}{y} e^{\frac{y}{x}}

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