What is the sum from n=1 to infinity of (3/2)^{3-2n} ?

The sum from n=1 to infinity of (3/2)^{3-2n} 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 (3/2)^{3-2n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 6 (\frac{8}{9})^{n}

n = 0 \sum \infty 0.5

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

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

k = 2 \sum \infty k e^{- 4 k^{2}}

What is the sum from n=1 to infinity of (3/2)^{3-2n} ?
The sum from n=1 to infinity of (3/2)^{3-2n} is converges