What is the sum from n=1 to infinity of (-1^n)/(n!) ?

The sum from n=1 to infinity of (-1^n)/(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 (-1^n)/(n!)

n = 1 \sum \infty \frac{- 1 ^{n}}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{- 1 ^{n}}{n !}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 1 \sum \infty \frac{1 ^{n}}{n !}

Apply Series Ratio Test:

converges

= - converges

= converges

x^{8 c o s (x)}

x \to 1 - lim (1 - x^{3})

\int \frac{2}{16 - 25 x ^{2}} d x

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

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

What is the sum from n=1 to infinity of (-1^n)/(n!) ?
The sum from n=1 to infinity of (-1^n)/(n!) is converges