What is the sum from n=1 to infinity of n!e^{-3n} ?

The sum from n=1 to infinity of n!e^{-3n} is diverges

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sum from n=1 to infinity of n!e^{-3n}

n = 1 \sum \infty n! e^{- 3 n}

diverges

Solution steps

n = 1 \sum \infty n! e^{- 3 n}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty cos (3 n)

n = 1 \sum \infty 2 n + i

n = 0 \sum \infty cos (\frac{2}{n})

n = 1 \sum \infty \frac{1}{n ^{6} + 8}

n = 1 \sum \infty \frac{5 n}{2 ^{n}}

What is the sum from n=1 to infinity of n!e^{-3n} ?
The sum from n=1 to infinity of n!e^{-3n} is diverges