What is the sum from n=1 to infinity of (n!)/(109^n) ?

The sum from n=1 to infinity of (n!)/(109^n) is diverges

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sum from n=1 to infinity of (n!)/(109^n)

n = 1 \sum \infty \frac{n !}{10 9 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{10 9 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

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

n = 1 \sum \infty \frac{2022}{9 ^{n}}

n = 1 \sum \infty n^{- 4}

n = 1 \sum \infty (1)^{n - 1}

n = 1 \sum \infty (2)^{n}

What is the sum from n=1 to infinity of (n!)/(109^n) ?
The sum from n=1 to infinity of (n!)/(109^n) is diverges