What is the sum from n=0 to infinity of (n!)/(9^n) ?

The sum from n=0 to infinity of (n!)/(9^n) is diverges

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sum from n=0 to infinity of (n!)/(9^n)

n = 0 \sum \infty \frac{n !}{9 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{n !}{9 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{9}{n ^{2} + 16}

n = 0 \sum \infty \frac{1}{ln ( n ^{n} )}

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

n = 1 \sum \infty \frac{ln ( 7 n )}{n}

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

What is the sum from n=0 to infinity of (n!)/(9^n) ?
The sum from n=0 to infinity of (n!)/(9^n) is diverges