What is the sum from n=1 to infinity of n^2*e^{-n} ?

The sum from n=1 to infinity of n^2*e^{-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^2*e^{-n}

n = 1 \sum \infty n^{2} \cdot e^{- n}

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\int_{1}^{2} 7 4 x^{2} - 3 d x

\int x^{- 10} d x

x \to 2 lim (\frac{- 1}{x - 2})

3 t sin (3 t)

\frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} = 0

What is the sum from n=1 to infinity of n^2*e^{-n} ?
The sum from n=1 to infinity of n^2*e^{-n} is converges