What is the sum from n=1 to infinity of (0.73)^{n-1} ?

The sum from n=1 to infinity of (0.73)^{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 (0.73)^{n-1}

n = 1 \sum \infty (0.73)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 0.7 3^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 2 \sum \infty 1^{l n (n)}

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

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

k = 0 \sum \infty (\frac{1}{7})^{k}

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

What is the sum from n=1 to infinity of (0.73)^{n-1} ?
The sum from n=1 to infinity of (0.73)^{n-1} is converges