What is the sum from n=1 to infinity of ((8pi)/9)^n ?

The sum from n=1 to infinity of ((8pi)/9)^n 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 ((8pi)/9)^n

n = 1 \sum \infty (\frac{8 π}{9})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{8 π}{9})^{n}

Apply Series Ratio Test:

diverges

= diverges

n = 2 \sum \infty \frac{n !}{6 ^{n}}

k = 1 \sum \infty \frac{6}{k ( k + 2 )}

k = 1 \sum \infty \frac{3}{k}

n = 3 \sum \infty \frac{1}{n ^{2} ln ( n )}

n = 2 \sum \infty \frac{5}{n ln ( n )}

What is the sum from n=1 to infinity of ((8pi)/9)^n ?
The sum from n=1 to infinity of ((8pi)/9)^n is diverges