What is the sum from n=1 to infinity of n!(-e)^{-4n} ?

The sum from n=1 to infinity of n!(-e)^{-4n} is diverges

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sum from n=1 to infinity of n!(-e)^{-4n}

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{6}{n ^{2} + 1}

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

n = 1 \sum \infty \frac{19}{1 + 3 ^{n}}

n = 1 \sum \infty \frac{ln ( n )}{7 n}

n = 4 \sum \infty \frac{5}{3 ^{n}}

What is the sum from n=1 to infinity of n!(-e)^{-4n} ?
The sum from n=1 to infinity of n!(-e)^{-4n} is diverges