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

The sum from n=1 to infinity of (1/8)^{n-1} is converges

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 (1/8)^{n-1}

n = 1 \sum \infty (\frac{1}{8})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{1}{8})^{n - 1}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (\frac{c e ^{x}}{1 + c e ^{x}})

y = x^{2} + 7, (3, 16)

\frac{d}{d x} (x^{3} + 2 x + 1)

\int sin (3 x + 2) d x

y = 4 x cos (x), at x = π

What is the sum from n=1 to infinity of (1/8)^{n-1} ?
The sum from n=1 to infinity of (1/8)^{n-1} is converges