What is the sum from n=1 to infinity of (n-3)/(2^n) ?

The sum from n=1 to infinity of (n-3)/(2^n) is converges

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sum from n=1 to infinity of (n-3)/(2^n)

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

converges

Solution steps

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

Expand

\frac{n - 3}{2 ^{n}} : \frac{n}{2 ^{n}} - \frac{3}{2 ^{n}}

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

Apply the Sum Rule:

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

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

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

converges

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

= converges - 3

= converges

n = 4 \sum \infty \frac{2}{5 n ^{3}}

n = 0 \sum \infty 1 + \frac{( - 1 ) ^{n}}{n}

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

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

k = 1 \sum \infty \frac{2}{k ln ( 9 k )}

What is the sum from n=1 to infinity of (n-3)/(2^n) ?
The sum from n=1 to infinity of (n-3)/(2^n) is converges