What is the sum from n=7 to infinity of (n!)/(n-6) ?

The sum from n=7 to infinity of (n!)/(n-6) is diverges

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sum from n=7 to infinity of (n!)/(n-6)

n = 7 \sum \infty \frac{n !}{n - 6}

diverges

Solution steps

n = 7 \sum \infty \frac{n !}{n - 6}

Apply Series Ratio Test:

diverges

= diverges

k = 1 \sum \infty \frac{1}{k ^{2} + 3 k + 2}

n = 2 \sum \infty \frac{10}{ln ( n )}

n = 0 \sum \infty \frac{9}{2} (\frac{1}{2})^{n}

n = 0 \sum \infty \frac{n ^{2} + 1}{n}

k = 0 \sum \infty (\frac{1}{2})^{k}

What is the sum from n=7 to infinity of (n!)/(n-6) ?
The sum from n=7 to infinity of (n!)/(n-6) is diverges