What is the sum from n=0 to infinity of (n!)/(2^n) ?

The sum from n=0 to infinity of (n!)/(2^n) is diverges

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sum from n=0 to infinity of (n!)/(2^n)

n = 0 \sum \infty \frac{n !}{2 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{n !}{2 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

x \to - \infty lim (x^{2} e x)

\frac{\partial}{\partial y} (4 - x^{2} - 2 y^{2})

\frac{d}{d x} (\frac{x - 2 ln ( x ) - 1}{x - 1})

x \to 0 lim ((x + \frac{1}{x}) \cdot e^{- \frac{1}{x ^{2}}})

y = 4 x^{2} - 3

What is the sum from n=0 to infinity of (n!)/(2^n) ?
The sum from n=0 to infinity of (n!)/(2^n) is diverges