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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

y = 18 sin (x), y = 18 cos (x)

x^{2} - 4 x + 3, - x^{2} + 2 x + 3, 4, 0

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 8

y = \frac{ln ( 6 x )}{6 x}

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