What is the sum from n=0 to infinity of (n!)/(3^n) ?

The sum from n=0 to infinity of (n!)/(3^n) is diverges

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sum from n=0 to infinity of (n!)/(3^n)

n = 0 \sum \infty \frac{n !}{3 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{n !}{3 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

\int 103 t^{2} d t

\frac{d}{d x} (9 x + \frac{1}{x} - 4)

x \to 3 lim (4 - x)

y = \frac{1}{( x ^{2} - 2 ) ( x ^{2} + x + 5 )}

x \to (\frac{π}{2}) + lim (tan (x))

What is the sum from n=0 to infinity of (n!)/(3^n) ?
The sum from n=0 to infinity of (n!)/(3^n) is diverges