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

The sum from n=1 to infinity of (-1)^{n-1} is diverges

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-1}

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial x} (2 x y - x^{3} - y^{2})

\int t (e^{- s t}) d t

\frac{3}{2} t + 2 e^{- t} - \frac{1}{4} e^{- 2 t} - \frac{7}{4}

\int \frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{( y + z )}{x})

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