What is the sum from n=1 to infinity of 0.9^nn^3 ?

The sum from n=1 to infinity of 0.9^nn^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 0.9^nn^3

n = 1 \sum \infty 0. 9^{n} n^{3}

converges

Solution steps

n = 1 \sum \infty 0. 9^{n} n^{3}

Apply Series Ratio Test:

converges

= converges

\int_{10}^{13} 2 x d x

y = x^{2} e^{- x}

\frac{d y}{d x} + \frac{y}{x + 2} = 5 (x - 2) y^{\frac{1}{2}}

y = 4 x - 3 x^{2}

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

What is the sum from n=1 to infinity of 0.9^nn^3 ?
The sum from n=1 to infinity of 0.9^nn^3 is converges