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

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

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

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

converges

Solution steps

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

Expand

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

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

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

converges

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

converges

= converges + converges

= converges

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