What is the sum from n=1 to infinity of (n!)/(94^n) ?

The sum from n=1 to infinity of (n!)/(94^n) is diverges

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sum from n=1 to infinity of (n!)/(94^n)

n = 1 \sum \infty \frac{n !}{9 4 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{9 4 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty 3 + (- 1)^{n}

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

n = 1 \sum \infty \frac{33 n !}{n ^{n}}

x = 0 \sum \infty x^{2}

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

What is the sum from n=1 to infinity of (n!)/(94^n) ?
The sum from n=1 to infinity of (n!)/(94^n) is diverges