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

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

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

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

converges

Solution steps

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

Simplify

2^{n - 1} : 2^{n} \cdot 2^{- 1}

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

\frac{1}{2 ^{- 1}} = 2

= 2 \cdot n = 2 \sum \infty \frac{n}{2 ^{n}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\frac{5}{( s ^{2} + 2 s + 2 )}

\frac{\partial}{\partial x} (\frac{( 4 x - y )}{4 x + y})

\frac{d}{d x} (100 x e^{- 0.5 x})

\frac{\partial}{\partial y} (6 x)

f (x) = x - e^{- x}

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