What is the sum from n=1 to infinity of (9^n+n)/(n!) ?

The sum from n=1 to infinity of (9^n+n)/(n!) is converges

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sum from n=1 to infinity of (9^n+n)/(n!)

n = 1 \sum \infty \frac{9 ^{n} + n}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{9 ^{n} + n}{n !}

Expand

\frac{9 ^{n} + n}{n !} : \frac{9 ^{n}}{n !} + \frac{n}{n !}

= n = 1 \sum \infty \frac{9 ^{n}}{n !} + \frac{n}{n !}

Apply the Sum Rule:

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

= n = 1 \sum \infty \frac{9 ^{n}}{n !} + n = 1 \sum \infty \frac{n}{n !}

n = 1 \sum \infty \frac{9 ^{n}}{n !} =

converges

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

converges

= converges + converges

= converges

n = 0 \sum \infty e^{5 n}

n = 1 \sum \infty cos (n + 1)

n = 0 \sum \infty \frac{3}{2 ^{2 n}}

n = 1 \sum \infty (\frac{12}{8})^{n}

n = 1 \sum \infty (cos (19))^{n}

What is the sum from n=1 to infinity of (9^n+n)/(n!) ?
The sum from n=1 to infinity of (9^n+n)/(n!) is converges