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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

(x, y) \to (1, 2) lim (1 + x + y)

f (x) = x^{3} - 5 + 2 x

x \to \infty lim (e^{- x})

\frac{\partial}{\partial x} (\frac{2 x - 3}{x y})

7 - 3 u

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