What is the sum from n=1 to infinity of 6/(n!) ?

The sum from n=1 to infinity of 6/(n!) is converges

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sum from n=1 to infinity of 6/(n!)

n = 1 \sum \infty \frac{6}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{6}{n !}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n !}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 1 \sum \infty (\frac{5}{4})^{n}

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

n = 1 \sum \infty \frac{2}{n ^{2} + 4}

n = 1 \sum \infty \frac{1}{n ( n + 4 )}

n = 1 \sum \infty \frac{e ^{n}}{n ^{8}}

What is the sum from n=1 to infinity of 6/(n!) ?
The sum from n=1 to infinity of 6/(n!) is converges