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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{6 + e ^{- n}}

n = 2 \sum \infty \frac{2}{n ln ( n )}

n = 1 \sum \infty 5^{- 3 n}

n = 1 \sum \infty \frac{1}{e ^{4 n}}

n = 2 \sum \infty (- 0.55)^{n}

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