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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} (e^{yz})

y^{^{'}} = a \cdot e^{- 2 x} + b \cdot e^{2 x} + \frac{3 x}{4} - \frac{1}{4}

f (x) = e^{x} sin (x) + cos (x) e^{x}

y = arccos (x)

\frac{1}{8 s + 1}

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