What is the sum from n=2 to infinity of (3/8)^{3n} ?

The sum from n=2 to infinity of (3/8)^{3n} 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=2 to infinity of (3/8)^{3n}

n = 2 \sum \infty (\frac{3}{8})^{3 n}

converges

Solution steps

n = 2 \sum \infty (\frac{3}{8})^{3 n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (1 + \frac{6}{n})^{n}

n = 1 \sum \infty 9 (0.9)^{n - 1}

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

n = 0 \sum \infty \frac{2}{n ^{3} + 5}

n = 0 \sum \infty cos (\frac{1}{n + 1})

What is the sum from n=2 to infinity of (3/8)^{3n} ?
The sum from n=2 to infinity of (3/8)^{3n} is converges