What is the sum from n=1 to infinity of (n!)/(3^n) ?

The sum from n=1 to infinity of (n!)/(3^n) is diverges

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sum from n=1 to infinity of (n!)/(3^n)

n = 1 \sum \infty \frac{n !}{3 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{3 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

x \to 0 lim (\frac{1 0 ^{x} - 1 1 ^{x}}{x})

\int_{- 2}^{2} (2 - ∣ x ∣) d x

x \to 0 - lim (- 3 x^{2} + 2 x)

\int_{0}^{l} x sin^{2} (\frac{π x}{l}) d x

(5.1) (3.6)

What is the sum from n=1 to infinity of (n!)/(3^n) ?
The sum from n=1 to infinity of (n!)/(3^n) is diverges