What is the sum from n=0 to infinity of n*e^{-n^2} ?

The sum from n=0 to infinity of n*e^{-n^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=0 to infinity of n*e^{-n^2}

n = 0 \sum \infty n \cdot e^{- n^{2}}

converges

Solution steps

n = 0 \sum \infty n e^{- n^{2}}

Apply Series Ratio Test:

converges

= converges

y = 4 ln (x)

\int \frac{x ^{2} + 1}{x ^{2}} d x

\int \frac{4}{x - 2} d x

\int 3 (3 x + 4)^{5} d x

\int \frac{sin ( 2 x )}{45 + cos ^{2} ( x )} d x

What is the sum from n=0 to infinity of n*e^{-n^2} ?
The sum from n=0 to infinity of n*e^{-n^2} is converges