What is the sum from n=1 to infinity of (n+4)/(4^n) ?

The sum from n=1 to infinity of (n+4)/(4^n) is converges

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sum from n=1 to infinity of (n+4)/(4^n)

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

converges

Solution steps

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

Expand

\frac{n + 4}{4 ^{n}} : \frac{n}{4 ^{n}} + 4^{1 - n}

= n = 1 \sum \infty \frac{n}{4 ^{n}} + 4^{1 - n}

Apply the Sum Rule:

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

= n = 1 \sum \infty \frac{n}{4 ^{n}} + n = 1 \sum \infty 4^{1 - n}

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

converges

n = 1 \sum \infty 4^{1 - n} =

converges

= converges + converges

= converges

What is the sum from n=1 to infinity of (n+4)/(4^n) ?
The sum from n=1 to infinity of (n+4)/(4^n) is converges