What is the sum from n=1 to infinity of (n!)/(104^n) ?

The sum from n=1 to infinity of (n!)/(104^n) is diverges

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sum from n=1 to infinity of (n!)/(104^n)

n = 1 \sum \infty \frac{n !}{10 4 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{10 4 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

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

n = 1 \sum \infty (- 0.6)^{n}

n = 1 \sum \infty \frac{2}{n ^{3} + 5}

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{3}}

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

What is the sum from n=1 to infinity of (n!)/(104^n) ?
The sum from n=1 to infinity of (n!)/(104^n) is diverges