What is the sum from n=0 to infinity of (1/12)^n ?

The sum from n=0 to infinity of (1/12)^n is 12/11

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 (1/12)^n

n = 0 \sum \infty (\frac{1}{12})^{n}

\frac{12}{11}

Solution steps

n = 0 \sum \infty (\frac{1}{12})^{n}

Apply Series Geometric Test:

\frac{12}{11}

= \frac{12}{11}

n = 4 \sum \infty ln (\frac{n}{2})

n = 1 \sum \infty \frac{1}{9 + n ^{2}}

n = 1 \sum \infty \frac{1}{n ^{\frac{π}{4}}}

n = 0 \sum \infty (\frac{7}{10})^{n}

n = 1 \sum \infty \frac{1}{n ^{\frac{1}{6}}}

What is the sum from n=0 to infinity of (1/12)^n ?
The sum from n=0 to infinity of (1/12)^n is 12/11