What is the sum from n=0 to infinity of 1/(n2^n) ?

The sum from n=0 to infinity of 1/(n2^n) is converges

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sum from n=0 to infinity of 1/(n2^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty 9 ln (\frac{n}{n + 1})

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

n = 1 \sum \infty \frac{9}{n ^{\frac{3}{7}}}

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

n = 1 \sum \infty (- e)^{- n}

What is the sum from n=0 to infinity of 1/(n2^n) ?
The sum from n=0 to infinity of 1/(n2^n) is converges