What is the sum from k=1 to infinity of [(k/(2^k))] ?

The sum from k=1 to infinity of [(k/(2^k))] is converges

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sum from k=1 to infinity of [(k/(2^k))]

k = 1 \sum \infty [(\frac{k}{2 ^{k}})]

converges

Solution steps

k = 1 \sum \infty \frac{k}{2 ^{k}}

Apply Series Ratio Test:

converges

= converges

k = 0 \sum \infty - 1^{k}

n = 1 \sum \infty \frac{5}{n ^{2} + n ^{3}}

n = 0 \sum \infty e^{- 1 n}

n = 1 \sum \infty \frac{2 n}{7 n - 6}

n = 1 \sum \infty n + 3

What is the sum from k=1 to infinity of [(k/(2^k))] ?
The sum from k=1 to infinity of [(k/(2^k))] is converges