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

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

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

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

converges

Solution steps

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

Expand

\frac{n + 2}{n !} : \frac{n}{n !} + \frac{2}{n !}

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

Apply the Sum Rule:

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

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

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

converges

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

converges

= converges + converges

= converges

\frac{d N}{d t} = N (1 - 0.0004 N)

\int_{0}^{1} \frac{x}{x ^{2} + x + 1} d x

\frac{d x}{d t} + 5 t^{7} x^{9} + \frac{x}{t} = 0

x \to 0 - lim (\frac{ln ( x ^{2} )}{x})

\frac{x ^{2}}{1 - 2 x}

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