What is the sum from n=1 to infinity of (n!)/(8^n) ?

The sum from n=1 to infinity of (n!)/(8^n) is diverges

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sum from n=1 to infinity of (n!)/(8^n)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 3 \sum \infty n^{- 0.9999}

n = 2 \sum \infty (- 0.85)^{n}

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

n = 0 \sum \infty (- \frac{6}{7})^{n}

n = 1 \sum \infty 90 (0.1)^{n}

What is the sum from n=1 to infinity of (n!)/(8^n) ?
The sum from n=1 to infinity of (n!)/(8^n) is diverges