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

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

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

k = 1 \sum \infty \frac{k !}{2 ^{k}}

diverges

Solution steps

k = 1 \sum \infty \frac{k !}{2 ^{k}}

Apply Series Ratio Test:

diverges

= diverges

\int 15 tan^{2} (x) sec^{3} (x) d x

d er i v a t i v e 2 x e^{4 x}

\int \frac{2}{x ^{3} x ^{2} - 3} d x

d er i v a t i v e 2 x + 3 x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x}

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