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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty \frac{e ^{\frac{3}{n}}}{n}

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

k = 1 \sum \infty \frac{1}{k ^{3}}

n = 1 \sum \infty \frac{6}{n + 3}

n = 2 \sum \infty (- 0.65)^{n}

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