What is the sum from n=1 to infinity of n!e^{-8n} ?

The sum from n=1 to infinity of n!e^{-8n} is diverges

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sum from n=1 to infinity of n!e^{-8n}

n = 1 \sum \infty n! e^{- 8 n}

diverges

Solution steps

n = 1 \sum \infty n! e^{- 8 n}

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n + sin ( n )}

n = 3 \sum \infty \frac{4}{ln ( n )}

n = 0 \sum \infty \frac{1}{7}

n = 0 \sum \infty \frac{1}{8}

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

What is the sum from n=1 to infinity of n!e^{-8n} ?
The sum from n=1 to infinity of n!e^{-8n} is diverges