What is the sum from n=1 to infinity of (n!)/(4^n) ?

The sum from n=1 to infinity of (n!)/(4^n) is diverges

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sum from n=1 to infinity of (n!)/(4^n)

n = 1 \sum \infty \frac{n !}{4 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{4 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

\int (x + 1) x d x

\frac{d}{d x} (\frac{x}{x})

\frac{\partial}{\partial x} (35 e^{5 x} y)

\int x^{\frac{2}{7}} d x

x^{2} - 2 x, at x = 2

What is the sum from n=1 to infinity of (n!)/(4^n) ?
The sum from n=1 to infinity of (n!)/(4^n) is diverges