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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

k = 1 \sum \infty \frac{4}{k} - \frac{4}{k + 1}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 8}

k = 6 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

n = 2 \sum \infty \frac{1}{4 n ^{3}}

n = 2 \sum \infty \frac{2 ^{n}}{n !}

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