What is the sum from n=0 to infinity of ((1/2)^n)^2 ?

The sum from n=0 to infinity of ((1/2)^n)^2 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=0 to infinity of ((1/2)^n)^2

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty 5 (\frac{1}{5})^{n - 1}

n = 1 \sum \infty \frac{ln ( n )}{9 n}

n = 2 \sum \infty \frac{2}{7 ^{n}}

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

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

What is the sum from n=0 to infinity of ((1/2)^n)^2 ?
The sum from n=0 to infinity of ((1/2)^n)^2 is converges