What is the sum from n=1 to infinity of (n+5)/(5^n) ?

The sum from n=1 to infinity of (n+5)/(5^n) is converges

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sum from n=1 to infinity of (n+5)/(5^n)

n = 1 \sum \infty \frac{n + 5}{5 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n + 5}{5 ^{n}}

Expand

\frac{n + 5}{5 ^{n}} : \frac{n}{5 ^{n}} + 5^{1 - n}

= n = 1 \sum \infty \frac{n}{5 ^{n}} + 5^{1 - n}

Apply the Sum Rule:

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

= n = 1 \sum \infty \frac{n}{5 ^{n}} + n = 1 \sum \infty 5^{1 - n}

n = 1 \sum \infty \frac{n}{5 ^{n}} =

converges

n = 1 \sum \infty 5^{1 - n} =

converges

= converges + converges

= converges

\int_{0}^{1} \frac{x ^{6}}{x ^{7} - 1} d x

\int \frac{3}{1 + x ^{2}} d x

y^{^{''}} + y = 4 x sin (x)

x \to + 3 lim ((\frac{x - 1}{2 x - 4})^{\frac{1}{x - 3}})

y^{^{''}} = - 4 \cdot y

What is the sum from n=1 to infinity of (n+5)/(5^n) ?
The sum from n=1 to infinity of (n+5)/(5^n) is converges