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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= 9 converges

= converges

(ln^{2} (x))^{^{'}}

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

\frac{d}{d x} (- 4 x)

\int 2 x (x^{2} + 2)^{6} d x

\frac{\partial}{\partial x} (3 x^{4} y^{2} - 2)

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