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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (\frac{1}{16 - x ^{2} - 5 x})

\int_{2}^{5} (\frac{3 x + 3}{x ^{2}}) d x

\int sin (5 x) cos (8 x) d x

\int x (x - 1) d x

\int_{1}^{\infty} \frac{n}{n ^{2} + 1}

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