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

\int \frac{x ^{2}}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = - \frac{2}{x}

\int \frac{x}{( x ^{2} + 4 ) ^{3}} d x

\int \frac{3}{x ^{2} 1 + x ^{2}} d x

\frac{\partial}{\partial y} (x^{2} + y^{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