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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 4 \sum \infty n e^{- n^{2}}

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

n = 0 \sum \infty cos (\frac{nπ}{2})

n = 0 \sum \infty [7 (0.1)^{n}]

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n + 2}

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