What is the sum from n=0 to infinity of n!e^{-2n} ?

The sum from n=0 to infinity of n!e^{-2n} 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=0 to infinity of n!e^{-2n}

n = 0 \sum \infty n! e^{- 2 n}

diverges

Solution steps

n = 0 \sum \infty n! e^{- 2 n}

Apply Series Ratio Test:

diverges

= diverges

\frac{3 s + 1}{s ( s + 1 ) ( - s + 5 )}

\int_{0}^{1} \frac{1}{x - 3} d x

y^{^{''}} - 81 y = 0, y (2) = 2, y (- 1) = - 1

\frac{d}{d x} (x + e)

\frac{1}{sin ( x ) cos ( x + b + c )}

What is the sum from n=0 to infinity of n!e^{-2n} ?
The sum from n=0 to infinity of n!e^{-2n} is diverges