What is the sum from n=0 to infinity of (3n!)/(n^2) ?

The sum from n=0 to infinity of (3n!)/(n^2) is diverges

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sum from n=0 to infinity of (3n!)/(n^2)

n = 0 \sum \infty \frac{3 n !}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n !}{n ^{2}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n !}{n ^{2}}

Apply Series Ratio Test:

diverges

= 3 diverges

= diverges

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

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

n = 0 \sum \infty \frac{5 ^{n}}{n ^{2}}

n = 0 \sum \infty \frac{8}{( 1 - 7 n )}

n = 1 \sum \infty 53 (\frac{1}{2})^{n}

What is the sum from n=0 to infinity of (3n!)/(n^2) ?
The sum from n=0 to infinity of (3n!)/(n^2) is diverges