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

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

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

converges

Solution steps

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

n^{- 1} = \frac{1}{n}

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

Apply Alternating Series Test:

converges

= converges

y^{^{''}} + 5 y^{^{'}} = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'}} = 8 x + y, y (0) = 6

\frac{\partial}{\partial z} (e^{x} + yz)

f (x)

x \to 6 + lim (\frac{x}{36 - x ^{2}})

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