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

The sum from n=0 to infinity of (1/pi)^n is pi/(pi-1)

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/pi)^n

n = 0 \sum \infty (\frac{1}{π})^{n}

\frac{π}{π - 1}

Solution steps

n = 0 \sum \infty (\frac{1}{π})^{n}

Apply Series Geometric Test:

\frac{π}{π - 1}

= \frac{π}{π - 1}

n = 0 \sum \infty 2 (\frac{3}{4})^{n}

n = 1 \sum \infty n e^{- 7 n^{2}}

k = 0 \sum \infty (0.6)^{k}

n = 1 \sum \infty \frac{n ^{2}}{7 ^{n}}

n = 0 \sum \infty \frac{4}{2 n + 3}

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