What is the sum from n=1 to infinity of n^3e^{-n} ?

The sum from n=1 to infinity of n^3e^{-n} is converges

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^3e^{-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (\frac{- 1 - x}{( x - 1 ) ^{3}})

\int \frac{( x ^{2} + x + 1 )}{x ^{2} + 1} d x

\frac{d}{d x} (ln (e^{4}))

\frac{d}{d x} (5 x^{3} - x^{- \frac{2}{5}})

x \to 0 lim (12 + x^{2} ln (x))

What is the sum from n=1 to infinity of n^3e^{-n} ?
The sum from n=1 to infinity of n^3e^{-n} is converges