What is the sum from n=1 to infinity of (n+3)/(8^n) ?

The sum from n=1 to infinity of (n+3)/(8^n) is converges

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sum from n=1 to infinity of (n+3)/(8^n)

n = 1 \sum \infty \frac{n + 3}{8 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n + 3}{8 ^{n}}

Expand

\frac{n + 3}{8 ^{n}} : \frac{n}{8 ^{n}} + \frac{3}{8 ^{n}}

= n = 1 \sum \infty \frac{n}{8 ^{n}} + \frac{3}{8 ^{n}}

Apply the Sum Rule:

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

= n = 1 \sum \infty \frac{n}{8 ^{n}} + n = 1 \sum \infty \frac{3}{8 ^{n}}

n = 1 \sum \infty \frac{n}{8 ^{n}} =

converges

n = 1 \sum \infty \frac{3}{8 ^{n}} = \frac{3}{7}

= converges + \frac{3}{7}

= converges

n = 0 \sum \infty \frac{k}{2 ^{n}}

n = 2 \sum \infty 0.3 3^{n}

n = 3 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

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

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

What is the sum from n=1 to infinity of (n+3)/(8^n) ?
The sum from n=1 to infinity of (n+3)/(8^n) is converges