What is the sum from n=3 to infinity of (n!)/(2^{n+1)} ?

The sum from n=3 to infinity of (n!)/(2^{n+1)} is diverges

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sum from n=3 to infinity of (n!)/(2^{n+1)}

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

\frac{\partial}{\partial y} (- \frac{y ^{2}}{( x - y ) ^{2}})

\int 7 d x

x \to - \infty lim (\frac{- 3 x ^{3} - 4 x ^{2} + 2}{5 x ^{3} + 4 x ^{2} - 2 x})

\frac{d}{d x} (e^{- x} (a cos (x) + b sin (x)))

What is the sum from n=3 to infinity of (n!)/(2^{n+1)} ?
The sum from n=3 to infinity of (n!)/(2^{n+1)} is diverges