What is the sum from n=0 to infinity of (n+1)/(2^nn) ?

The sum from n=0 to infinity of (n+1)/(2^nn) is converges

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sum from n=0 to infinity of (n+1)/(2^nn)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{7} + 7}

n = 1 \sum \infty \frac{1}{n ^{7} + 2}

k = 0 \sum \infty \frac{k}{1 + k ^{4}}

n = 1 \sum \infty \frac{5}{n ^{0.6}}

n = 1 \sum \infty \frac{1}{4 n + 5}

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