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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{4}{n + 3}

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

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

n = 0 \sum \infty sin (\frac{nπ}{2})

n = 1 \sum \infty \frac{1}{n ^{2} + 4 n}

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