What is the sum from n=1 to infinity of n!e^{-9n} ?

The sum from n=1 to infinity of n!e^{-9n} 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^{-9n}

n = 1 \sum \infty n! e^{- 9 n}

diverges

Solution steps

n = 1 \sum \infty n! e^{- 9 n}

Apply Series Ratio Test:

diverges

= diverges

n = 1 \sum \infty (3.5 + 1)^{n}

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

n = 500 \sum \infty \frac{1}{n}

n = 2 \sum \infty \frac{ln ( n )}{n}

x = 0 \sum \infty \frac{x}{1 - 4 x}

What is the sum from n=1 to infinity of n!e^{-9n} ?
The sum from n=1 to infinity of n!e^{-9n} is diverges