What is the sum from n=0 to infinity of (n!)/(4^n) ?

The sum from n=0 to infinity of (n!)/(4^n) is diverges

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sum from n=0 to infinity of (n!)/(4^n)

n = 0 \sum \infty \frac{n !}{4 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{n !}{4 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

\int_{8}^{9} \frac{x}{x - 8} d x

\frac{\partial}{\partial x} (\frac{2 x}{1 + x + 2 y})

f (x) = ln (2 x + 1)

\frac{e ^{x}}{8 x ^{2} + 1}

\int \frac{x}{( 2 + x ^{\frac{3}{2}} ) ^{3}} d x

What is the sum from n=0 to infinity of (n!)/(4^n) ?
The sum from n=0 to infinity of (n!)/(4^n) is diverges