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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{n}{n !}

Apply Series Ratio Test:

converges

= 5 converges

= converges

\frac{d}{d x} ((\frac{x}{5} - \frac{5}{x})^{5})

\int_{0}^{2} (2 x - 3) (4 x^{2} + 1) d x

f (x) = 4 x^{2} + 4 x - 2, at x = - 1

\frac{d}{d x} ((x^{3} + 5) e^{x})

\int cos (4 x) 2 - sin (4 x) d x

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