What is the sum from n=1 to infinity of (n!)/(84^n) ?

The sum from n=1 to infinity of (n!)/(84^n) is diverges

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sum from n=1 to infinity of (n!)/(84^n)

n = 1 \sum \infty \frac{n !}{8 4 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{8 4 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

n = 2 \sum \infty \frac{4}{ln ( n ^{2} )}

n = 0 \sum \infty \frac{1}{1 + 8 n}

k = 1 \sum \infty \frac{1}{8 ^{k} + 4}

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

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

What is the sum from n=1 to infinity of (n!)/(84^n) ?
The sum from n=1 to infinity of (n!)/(84^n) is diverges