What is the sum from n=1 to infinity of (3n)/(n!) ?

The sum from n=1 to infinity of (3n)/(n!) is converges

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sum from n=1 to infinity of (3n)/(n!)

n = 1 \sum \infty \frac{3 n}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{3 n}{n !}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{n !}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int csc^{3} (x) \cdot sec (x) d x

\int (x^{2} + a x - 3)^{2} d x

\int_{0}^{1} 2 x^{2} - 1 d x

\frac{d ^{2}}{d x ^{2}} ((x + 1) e^{x})

n = 1 \sum \infty sin (πn)

What is the sum from n=1 to infinity of (3n)/(n!) ?
The sum from n=1 to infinity of (3n)/(n!) is converges