What is the sum from n=1 to infinity of 9n^2e^{-n^3} ?

The sum from n=1 to infinity of 9n^2e^{-n^3} 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 9n^2e^{-n^3}

n = 1 \sum \infty 9 n^{2} e^{- n^{3}}

converges

Solution steps

n = 1 \sum \infty 9 n^{2} e^{- n^{3}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 9 \cdot n = 1 \sum \infty n^{2} e^{- n^{3}}

Apply Series Ratio Test:

converges

= 9 converges

= converges

n = 0 \sum \infty (1 + \frac{1}{2 n})^{n}

n = 1 \sum \infty ln (\frac{3 n + 1}{n})

n = \frac{π}{2} \sum \infty sin (n)

n = 1 \sum \infty \frac{3}{( n ( n + 3 ) )}

k = 1 \sum \infty \frac{9}{k ( k + 2 )}

What is the sum from n=1 to infinity of 9n^2e^{-n^3} ?
The sum from n=1 to infinity of 9n^2e^{-n^3} is converges