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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{cos ( πn )}{5}

n = 11 \sum \infty \frac{17}{3 ^{n}}

n = 0 \sum \infty sec (\frac{1}{n ^{2}})

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

n = 1 \sum \infty \frac{3}{8 ^{n} + 1}

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