What is the sum from n=1 to infinity of n!e^{-2n} ?

The sum from n=1 to infinity of n!e^{-2n} is diverges

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sum from n=1 to infinity of n!e^{-2n}

n = 1 \sum \infty n! e^{- 2 n}

diverges

Solution steps

n = 1 \sum \infty n! e^{- 2 n}

Apply Series Ratio Test:

diverges

= diverges

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

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

n = 1 \sum \infty \frac{4 n}{n ^{2} + 5}

n = 0 \sum \infty (\frac{5}{6})^{n - 1}

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

What is the sum from n=1 to infinity of n!e^{-2n} ?
The sum from n=1 to infinity of n!e^{-2n} is diverges