What is the sum from n=0 to infinity of (n!)^2(0.5) ?

The sum from n=0 to infinity of (n!)^2(0.5) is diverges

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sum from n=0 to infinity of (n!)^2(0.5)

n = 0 \sum \infty (n!)^{2} (0.5)

diverges

Solution steps

n = 0 \sum \infty n!^{2} \cdot 0.5

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 0.5 \cdot n = 0 \sum \infty n!^{2}

Apply Series Ratio Test:

diverges

= 0.5 diverges

= diverges

k = 0 \sum \infty (- \frac{2}{5})^{k}

n = 1 \sum \infty 8 n + 1

n = 0 \sum \infty sin (e^{- n})

n = 1 \sum \infty 2 (- \frac{1}{7})^{2 n}

n = 1 \sum \infty \frac{1 9 ^{n}}{n !}

What is the sum from n=0 to infinity of (n!)^2(0.5) ?
The sum from n=0 to infinity of (n!)^2(0.5) is diverges