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

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

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

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

converges

Solution steps

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

Expand

\frac{n - 1}{n \cdot 2 ^{n}} : \frac{1}{2 ^{n}} - \frac{1}{2 ^{n} n}

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

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

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

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

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

converges

= 1 - converges

= converges

n = 1 \sum \infty 9^{- n}

n = 2 \sum \infty (\frac{1}{4})^{n}

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

n = 0 \sum \infty \frac{3 ^{n}}{e ^{n}}

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

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