What is the sum from n=1 to infinity of (n!)/(7^n) ?

The sum from n=1 to infinity of (n!)/(7^n) is diverges

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sum from n=1 to infinity of (n!)/(7^n)

n = 1 \sum \infty \frac{n !}{7 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{7 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{( 3 n - 1 ) ^{2}}

k = 1 \sum \infty k

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

n = 2 \sum \infty (- 0.25)^{n}

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

What is the sum from n=1 to infinity of (n!)/(7^n) ?
The sum from n=1 to infinity of (n!)/(7^n) is diverges