What is the sum from n=1 to infinity of (n!)/(n*2^n) ?

The sum from n=1 to infinity of (n!)/(n*2^n) is diverges

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sum from n=1 to infinity of (n!)/(n*2^n)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{25 n ^{2} + 5 n - 6}

n = 0 \sum \infty (- 1)^{n} \frac{1}{n}

n = 1 \sum \infty 2 (- \frac{4}{3})^{n}

n = 2 \sum \infty \frac{1}{n ^{1} ln ( n )}

n = 2 \sum \infty \frac{3}{n ( n + 3 )}

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