What is the sum from n=0 to infinity of (n-1)/(n+1) ?

The sum from n=0 to infinity of (n-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=0 to infinity of (n-1)/(n+1)

n = 0 \sum \infty \frac{n - 1}{n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{n - 1}{n + 1}

Apply Series Divergence Test:

diverges

= diverges

x^{2} - 4 x + 3, x

(4 t + 2)^{2} (2 t^{2} - 9)^{- 3}

2 e^{x}

\int_{0}^{π} cos^{6} (θ) d θ

\frac{d}{d x} (- \frac{1}{7})

What is the sum from n=0 to infinity of (n-1)/(n+1) ?
The sum from n=0 to infinity of (n-1)/(n+1) is diverges