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

The sum from n=1 to infinity of 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=1 to infinity of e^{(-n^2)}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

x \to - 5 lim (- \frac{9}{x + 5})

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

\int (5 + x) d x

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

x \to 3 lim (4 x - a + b)

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