What is the sum from n=0 to infinity of (3n)/(n!) ?

The sum from n=0 to infinity of (3n)/(n!) is converges

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sum from n=0 to infinity of (3n)/(n!)

n = 0 \sum \infty \frac{3 n}{n !}

converges

Solution steps

n = 0 \sum \infty \frac{3 n}{n !}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{n}{n !}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 1 \sum \infty \frac{1}{( n ln ( n ) )}

n = 0 \sum \infty \frac{i}{n ( n + 1 )}

n = 1 \sum \infty \frac{5 ^{n}}{n ^{3}}

n = 1 \sum \infty \frac{6 e ^{\frac{1}{n}}}{n}

k = 1 \sum \infty 1

What is the sum from n=0 to infinity of (3n)/(n!) ?
The sum from n=0 to infinity of (3n)/(n!) is converges