What is the sum from n=0 to infinity of (-1)^ne^{-n} ?

The sum from n=0 to infinity of (-1)^ne^{-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=0 to infinity of (-1)^ne^{-n}

n = 0 \sum \infty (- 1)^{n} e^{- n}

converges

Solution steps

n = 0 \sum \infty (- 1)^{n} e^{- n}

Apply Alternating Series Test:

converges

= converges

\int e^{9 x} cos (x) d x

(\frac{( x + 1 )}{x - 1})^{3}

x \to 1 lim (1 - x)

\int \frac{3}{x ^{2}} cos (\frac{3}{x}) d x

\frac{1}{s ^{3} + 6 s ^{2} + 9 s}

What is the sum from n=0 to infinity of (-1)^ne^{-n} ?
The sum from n=0 to infinity of (-1)^ne^{-n} is converges