What is the sum from n=1 to infinity}5^{-n of n! ?

The sum from n=1 to infinity}5^{-n of n! is diverges

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sum from n=1 to infinity}5^{-n of n!

n = 1 \sum \infty 5^{- n} n!

diverges

Solution steps

n = 1 \sum \infty 5^{- n} n!

Simplify

5^{- n} : \frac{1}{5 ^{n}}

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

Apply Series Ratio Test:

diverges

= diverges

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

k = 0 \sum \infty (\frac{3}{5})^{k}

n = 1 \sum \infty 5 (\frac{1}{π})^{n}

n = 1 \sum \infty \frac{6}{n + 5}

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

What is the sum from n=1 to infinity}5^{-n of n! ?
The sum from n=1 to infinity}5^{-n of n! is diverges