What is the sum from n=1 to infinity of (n-8)/(n+8) ?

The sum from n=1 to infinity of (n-8)/(n+8) 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 (n-8)/(n+8)

n = 1 \sum \infty \frac{n - 8}{n + 8}

diverges

Solution steps

n = 1 \sum \infty \frac{n - 8}{n + 8}

Apply Series Divergence Test:

diverges

= diverges

\int (e^{x} + a^{x}) d x

\int \frac{1}{( 1 + x )} d x

\int_{0}^{π} sin (2 x) cos (x) d x

y^{^{''}} - y^{^{'}} = 3 x^{2} e^{x}

\frac{d}{d x} (\frac{1 + x}{2})

What is the sum from n=1 to infinity of (n-8)/(n+8) ?
The sum from n=1 to infinity of (n-8)/(n+8) is diverges