What is the sum from n=1 to infinity of n^32^{-n} ?

The sum from n=1 to infinity of n^32^{-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^32^{-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\int (\frac{1}{x ^{2}} + \frac{4}{x x} + 2) d x

\frac{d}{d x} (\frac{3 x}{x ^{2} - 1})

\frac{d}{d x} (\frac{5 x - 4}{3 x ^{2} + 1})

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = - 1, y^{^{'}} (0) = 13

f (x) = \frac{28}{x}

What is the sum from n=1 to infinity of n^32^{-n} ?
The sum from n=1 to infinity of n^32^{-n} is converges