What is the sum from n=1 to infinity of 1/(4^{n-1)} ?

The sum from n=1 to infinity of 1/(4^{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/(4^{n-1)}

n = 1 \sum \infty \frac{1}{4 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{4 ^{n - 1}}

Apply Series Ratio Test:

converges

= converges

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

n = 0 \sum \infty (\frac{1}{20})^{n}

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

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

n = 1 \sum \infty \frac{1}{( 4 n + 5 ) ^{3}}

What is the sum from n=1 to infinity of 1/(4^{n-1)} ?
The sum from n=1 to infinity of 1/(4^{n-1)} is converges