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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y = (2 x + e^{- x^{2}})^{3}

\frac{d P}{d t} = P (a - b P)

y = 2 x, y = 2 x^{2} - 4

\frac{d}{d x} (∣ 2 x + 1 ∣)

\int_{1}^{8} \frac{2}{x ^{3}} d x

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