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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 9 converges

= converges

n = 0 \sum \infty \frac{n}{5 ^{n}}

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

n = 0 \sum \infty \frac{n}{1 + 2 n ^{2}}

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

n = 0 \sum \infty \frac{1}{n ^{2} + 4}

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