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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

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